<![CDATA[Problem A: Shopping]]>

Problem A: Shopping

You have just moved into a new apartment and have a long list of items you need to buy. Unfortunately, to buy this many items requires going to many different stores. You would like to minimize the amount of driving necessary to buy all the items you need.

Your city is organized as a set of intersections connected by roads. Your house and every store is located at some intersection. Your task is to find the shortest route that begins at your house, visits all the stores that you need to shop at, and returns to your house.

]]> The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing two integers N and M, the number of intersections and roads in the city, respectively. Each of these integers is between 1 and 100000, inclusive. The intersections are numbered from 0 to N-1. Your house is at the intersection numbered 0. M lines follow, each containing three integers X, Y, and D, indicating that the intersections X and Y are connected by a bidirectional road of length D. The following line contains a single integer S, the number of stores you need to visit, which is between 1 and ten, inclusive. The subsequent S lines each contain one integer indicating the intersection at which each store is located. It is possible to reach all of the stores from your house.

]]> For each test case, output a line containing a single integer, the length of the shortest possible shopping trip from your house, visiting all the stores, and returning to your house.

]]> #include #include #include #include using namespace std; typedef pair piu; struct road { int s; int d; int l; }; road roads[202000]; int first[202000]; unsigned best[202000]; int comp(const void* vr1, const void* vr2) { const road* r1 = (const road*) vr1; const road* r2 = (const road*) vr2; int ret; ret = r1->s - r2->s; if(ret) return ret; ret = r1->d - r2->d; if(ret) return ret; ret = r1->l - r2->l; if(ret) return ret; return ret; } int dist(int s, int d) { memset(best,-1,sizeof(best)); queue q; q.push(piu(s,0)); while(q.size() > 0) { piu cl = q.front(); q.pop(); int c = cl.first; unsigned l = cl.second; if(l >= best[c]) continue; best[c] = l; if(l >= best[d]) continue; for(int i = first[c]; roads[i].s == c; i++) { q.push(piu(roads[i].d, l+roads[i].l)); } } if(best[d] == -1) *((int*)5) = 0; return best[d]; } int store[20]; int dists[20][20]; int perm[20]; int S; unsigned bestest; void print(const int *v, const int size) { unsigned ret = 0; int i; if (v != 0) { for (i = 0; i < size; i++) { int j; if(i == 0) j = S; else j = v[i-1]-1; 0&&printf("(%d,%d) ", j, v[i]-1); ret += dists[j][v[i]-1]; 0&&printf("d%d ", dists[j][v[i]-1]); 0&&printf("%d ", v[i]); } 0&&printf("(%d,%d) ", v[i-1]-1, S); ret += dists[v[i-1]-1][S]; 0&&printf("d%d ", dists[v[i-1]-1][S]); 0&&printf("%d\n", ret); if(ret < bestest) bestest = ret; } } // print void visit(int *Value, int N, int k) { int i; static int level = -1; level = level+1; Value[k] = level; if (level == N) print(Value, N); else for (i = 0; i < N; i++) if (Value[i] == 0) visit(Value, N, i); level = level-1; Value[k] = 0; } main() { int NN; scanf("%d", &NN); while(NN--) { int N, M; memset(roads, 0, sizeof(roads)); memset(first, 0, sizeof(first)); scanf("%d %d", &N, &M); for(int i = 0; i < M; i++) { scanf("%d %d %d", &roads[2*i].s, &roads[2*i].d, &roads[2*i].l); roads[2*i+1].s = roads[2*i].d; roads[2*i+1].d = roads[2*i].s; roads[2*i+1].l = roads[2*i].l; } qsort(roads, M*2, sizeof(road), comp); roads[M*2].s = -1; for(int i = 0; i < N; i++) { first[i] = M*2; } for(int i = 0; i < M*2; i++) { if(first[roads[i].s] == M*2) { first[roads[i].s] = i; } } scanf("%d", &S); for(int i = 0; i < S; i++) { scanf("%d", store+i); } store[S] = 0; for(int ii = 0; ii < S+1; ii++) { for(int jj = 0; jj < S+1; jj++) { int i = store[ii]; int j = store[jj]; dists[ii][jj] = dist(i,j); } } bestest = -1; for(int i = 0; i < S; i++) perm[i] = 0; visit(perm, S, 0); printf("%d\n", bestest); } } ]]> <![CDATA[Problem B: Stack Machine]]>

A mathematician observes one person get on a bus. Then two people get off the bus. The mathematician says: "If one more person gets on the bus, the bus will be empty."

A stack machine is a special kind of bus. It only has doors at the front, and it is so narrow that the people on the bus cannot move past each other. Of the people on the bus, the person who got on the bus last must be the first to get off.

The bus travels in a city in which roads connect intersections, and all the roads are unidirectional. Along each road, a person either gets on or off the bus.

The mathematician does not know the identity of the bus passengers, but can estimate their height. Note that there may be multiple people with the same height.

Your task is to plan the route of the bus. The bus must be empty at the beginning and end of the route. Along the route, it must pick up and drop off the people corresponding to the roads it travels on. The height of the person that gets on or off the bus is fixed for each road.

]]> The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing three integers N, M, Q, the number of intersections and roads in the city and the number of queries, respectively. The number of intersections is between 1 and 100, inclusive. The number of roads and the number of queries are each between 1 and 100000, inclusive. Intersections in the city are numbered from 1 to N. The first line of each test case is followed by M lines describing the roads. Each of these lines contains three integers X, Y, and Z. These integers indicate that a road exists from intersection X to intersection Y, and that when the bus travels on this road, a person who is Z centimetres tall gets on the bus, if Z is positive, or a person who is -Z centimetres tall gets off the bus, if Z is negative. For example, Z=170 indicates that a 170 cm tall person gets on the bus, and Z=-170 indicates that a 170 cm tall person gets off the bus. Each person is at least 40 cm and at most 220 cm tall. The lines describing the roads are followed by Q more lines, each describing a query. A line describing a query contains two integers, the beginning and ending intersections of the bus route.

]]> For each query, output a line containing a single integer giving the length of the shortest non-empty path that the bus can take from the beginning to the end of the route. The input data will be such that this length is no more than 10^9. If there is no such path, output a line containing the word impossible.

]]> #include #include #include #include #include #include using namespace std; typedef pair pi; struct road { int s; int d; int p; }; bool operator<(const road r1, const road r2) { if(r1.s < r2.s) return true; if(r1.s > r2.s) return false; if(r1.d < r2.d) return true; if(r1.d > r2.d) return false; if(r1.p < r2.p) return true; if(r1.p > r2.p) return false; return false; } queue q; bitset<220> inq[220]; int bridge[220][220]; int qs; int deqs; void addbridge(int s, int d, unsigned l) { if(l%2) *((int*)5)=0; unsigned cur = bridge[s][d]; if(cur <= l) return; bridge[s][d] = l; q.push(pi(s,d)); inq[s].set(d); qs++; } set from[220]; set to[220]; map > top[220]; main() { int NN; scanf("%d", &NN); while(NN--) { int N, M, Q; scanf("%d %d %d", &N, &M, &Q); for(int i = 0; i < N; i++) { from[i] = set(); to[i] = set(); top[i] = map >(); } memset(bridge, -1, sizeof(bridge)); for(int i = 0; i < M; i++) { int x,y,z; scanf("%d %d %d", &x, &y, &z); 0&&printf("R: %d %d %d\n", x, y, z); road r; x--; y--; r.s = x; r.d = y; r.p = z; from[x].insert(r); to[y].insert(r); top[y][z].insert(r); } q = queue(); for(int i = 0; i < N; i++) { set fromi = from[i]; for(set::iterator it = fromi.begin(); it != fromi.end(); it++) { int p = it->p; if(p<0) continue; int s = it->d; set froms = from[s]; for(set::iterator it2 = froms.begin(); it2 != froms.end(); it2++) { if(it2->p == -p) addbridge(i, it2->d, 2); } } } while(q.size() > 0) { pi sd = q.front(); int s = sd.first; int d = sd.second; q.pop(); if(!inq[s].test(d)) continue; inq[s].reset(d); deqs++; set fromd = from[d]; for(set::iterator it = fromd.begin(); it != fromd.end(); it++) { int p = it->p; if(p>0) continue; int dd = it->d; set topsp = top[s][-p]; for(set::iterator it2 = topsp.begin(); it2 != topsp.end(); it2++) { int ss = it2->s; addbridge(ss, dd, bridge[s][d]+2); } } for(int dd = 0; dd < N; dd++) { if(bridge[d][dd] != -1) { addbridge(s, dd, bridge[s][d]+bridge[d][dd]); } if(bridge[dd][s] != -1) { addbridge(dd, d, bridge[s][d]+bridge[dd][s]); } } } for(int i = 0; i < Q; i++) { int x,y; scanf("%d %d\n", &x, &y); 0&&printf("Q: %d %d\n", x, y); x--; y--; if(bridge[x][y] == -1) puts("impossible"); else printf("%d\n", bridge[x][y]); } 0&&printf("qs: %d deqs: %d\n", qs, deqs); } } ]]> <![CDATA[Problem C: Ideas]]> A unique feature of ideas is that they are not consumed when used. A good idea can benefit arbitrarily many people without diminishing the value of the idea. An idea can even serve as the basis for creative people to derive even better ideas. Each person relies on a specific set of ideas to build on and create new ideas.

In order to realize these benefits, ideas need to be communicated to the people who use them. People have developed an extensive worldwide communication network to satisfy this need. The network is composed of a series of tubes connecting people. The tubes are inhabited by curious creatures called packets which carry ideas from one person to another. In order to avoid collisions between packets, the tubes are all unidirectional. Each person can have zero or more incoming and zero or more outgoing tubes.

All of the packets start at the same person, and each packet follows the following algorithm:

Learn and remember the ideas created by the current person.
If there are no outgoing tubes from the current person, stop executing the algorithm.
Otherwise, choose an arbitrary outgoing tube and use it to travel to a new person.
Tell the new person all of the ideas that he or she needs from other people.
Go back to step 1.

The input data is such that it is possible for a packet to reach every person from person 0, and whenever a person P relies on a given idea, every path a packet could have taken to reach P will have visited at least one person who created that idea. It is possible for more than one person to independently create the same idea.

To ease the strain on each packet, you would like to minimize the number of ideas that it remembers at any given time by directing the packet to forget certain ideas in certain tubes. However, in so doing, you must ensure that no matter what path the packet takes, every time it visits a person, the packet knows all of the ideas that the person needs.

]]> The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing three integers N, M, I, the number of people, tubes, and ideas, respectively. Each of these integers is between 1 and one thousand, inclusive. People are numbered from 0 to N-1, ideas from 0 to I-1. All of the packets start at person 0. 2N lines follow, two lines for each person in order from 0 to N-1. Each of these lines contains some number of integers separated by spaces. For each person, the first line lists the ideas that the person needs, and the second line lists the ideas that the person creates. A given idea will never appear on both lines for the same person. M more lines follow, each describing a tube using two integers: the source and destination person of the tube.

]]> For each test case, output M lines, each corresponding to one of the tubes in the same order as in the input. For each tube, output a list of integers: the minimal set of ideas the packet must have in its memory when travelling through the given tube. Output the ideas in each set in increasing order.

]]> #include #include #include #include using namespace std; bitset<1010> pres[1010]; bitset<1010> use[1010]; bitset<1010> bef[1010]; bitset<1010> aft[1010]; set preds[1010]; char buf[100000]; bitset<1010> seen; vector po; int endofedge[1010]; int N, M, I; void printset(char* msg, bitset<1010> s) { bool first = true; printf(msg); for(int i = 0; i < I; i++) { if(s.test(i)) { if(!first) printf(" "); printf("%d", i); first = false; } } puts(""); } bitset<1010> readset() { bitset<1010> ret = bitset<1010>(); gets(buf); char* tok = strtok(buf, " "); while(tok) { int i; if(1 == sscanf(tok, "%d", &i)) { ret.set(i); } tok = strtok(0, " "); } return ret; } void dfs(int i) { if(seen[i]) return; seen.set(i); for(set::iterator it = preds[i].begin(); it != preds[i].end(); it++) { dfs(*it); } po.push_back(i); } main() { int NN; gets(buf); sscanf(buf, "%d ", &NN); while(NN--) { gets(buf); sscanf(buf, "%d %d %d ", &N, &M, &I); for(int i = 0; i < N; i++) { use[i] = readset(); pres[i] = ~readset(); bef[i] = bitset<1010>(); aft[i] = bitset<1010>(); preds[i] = set(); } for(int i = 0; i < M; i++) { int x,y; scanf("%d %d ", &x, &y); preds[y].insert(x); endofedge[i] = y; } seen = bitset<1010>(); po = vector(); for(int i = 0; i < N; i++) { dfs(i); } bool change = true; while(change) { change = false; for(vector::reverse_iterator it = po.rbegin(); it != po.rend(); it++) { bitset<1010> old = bef[*it]; bef[*it] = (aft[*it] & pres[*it]) | use[*it]; if(bef[*it] != old) change = true; for(set::iterator it2 = preds[*it].begin(); it2 != preds[*it].end(); it2++) { old = aft[*it2]; aft[*it2] |= bef[*it]; if(aft[*it2] != old) change = true; } } } for(int i = 0; i < M; i++) { printset("", bef[endofedge[i]]); } /* for(int i = 0; i < N; i++) { printf("%d:\n", i); printset("use " , use[i]); printset("pres ", pres[i]); printset("bef ", bef[i]); printset("aft ", aft[i]); } */ } } ]]> <![CDATA[Problem D: HST]]> On July 1st, Ontario's Provincial Sales Tax (PST) was merged with the Federal Goods and Services Tax (GST), creating the Harmonized Sales Tax (HST). This changed the rate of tax Ontarians pay on various items.

Dalton would like to calculate how this change will affect his personal monthly budget.

]]> The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing two integers N, M, the number of categories of items and the number of purchases Dalton makes each month, respectively. Each of these integers is between 1 and 100000, inclusive. N lines follow, each describing a category of item. Each of these lines contains a category name, which is a string of at most 30 uppercase letters, followed by three percentages, giving the PST, GST, and HST rate on that category. Each percentage is a number with up to two digits after the decimal point, and each percentage is followed by the % symbol. Each percentage is at least 0% and at most 100%. These lines are followed by M more lines, each describing one of Dalton's purchases. Each of these lines contains a category name and a price in dollars and cents and prefixed by the $ symbol. Each such line indicates the amount of pre-tax money that Dalton spends on an item in the specified category. The amount of each tax to be paid on an item is rounded to the nearest cent; if the tax amount is exactly half a cent more than a whole number of cents, it is rounded up to the nearest greater whole number of cents.

]]> For each test case, output the difference, in dollars and cents, between the total HST payable and the total sum of PST and GST payable on Dalton's monthly purchases. If the HST is more than the PST+GST, output a positive amount. If the HST is less than the PST+GST, output a negative amount.

]]> #include #include #define ll long long using namespace std; ll tax(ll cents, ll rate) { return (cents * rate + 5000) / 10000; } main() { int NN; scanf("%d", &NN); while(NN--) { map pst; map gst; map hst; int N, M; scanf("%d %d ", &N, &M); while(N--) { char name[100]; double pp, gg, hh; scanf("%s %lf%% %lf%% %lf%%\n", &name, &pp, &gg, &hh); ll p = pp*100+.5; ll g = gg*100+.5; ll h = hh*100+.5; pst[string(name)] = p; gst[string(name)] = g; hst[string(name)] = h; } ll cents = 0; while(M--) { char name[100]; double pp; scanf("%s $%lf\n", &name, &pp); ll p = pp*100+.5; string nm = string(name); cents += tax(p, hst[nm]) - tax(p, gst[nm]) - tax(p, pst[nm]); } printf("%.02lf\n", cents*.01); } } ]]> <![CDATA[Problem E: Tunnelling the Earth]]> There are different methods of transporting people from place to place: cars, bikes, boats, trains, planes, etc. For very long distances, people generally fly in a plane. But this has the disadvantage that the plane must fly around the curved surface of the earth. A distance travelled would be shorter if the traveller followed a straight line from one point to the other through a tunnel through the earth.

For example, travelling from Waterloo to Cairo requires a distance of 9293521 metres following the great circle route around the earth, but only 8491188 metres following the straight line through the earth.

For this problem, assume that the earth is a perfect sphere with radius of 6371009 metres.

]]> The first line of input contains a single integer, the number of test cases to follow. Each test case is one line containing four floating point numbers: the latitude and longitude of the origin of the trip, followed by the latitude and longitude of the destination of the trip. All of these measurements are in degrees. Positive numbers indicate North latitude and East longitude, while negative numbers indicate South latitude and West longitude.

]]> For each test case, output a line containing a single integer, the difference in the distance between the two points following the great circle route around the surface of the earth and following the straight line through the earth, in metres. Round the difference of the distances to the nearest integer number of metres.

]]> #include double r = 6371009; double sqr(double x) { return x*x; } double torad(double deg) { return deg/180*M_PI; } main() { int N; scanf("%d ", &N); while(N--) { double alat, alng, blat, blng; scanf("%lf %lf %lf %lf ", &alat, &alng, &blat, &blng); //printf("%lf %lf %lf %lf\n", alat, alng, blat, blng); alat = torad(alat); alng = torad(alng); blat = torad(blat); blng = torad(blng); double ax = cos(alat) * cos(alng); double ay = cos(alat) * sin(alng); double az = sin(alat); double bx = cos(blat) * cos(blng); double by = cos(blat) * sin(blng); double bz = sin(blat); double dist = r*sqrt(sqr(ax-bx)+ sqr(ay-by)+sqr(az-bz)); double arc = r * acos(ax*bx + ay*by + az*bz); //printf("%lf %lf\n", dist, arc); double TOL=0.00001; int a1 = arc-dist+TOL+.5; int a2 = arc-dist-TOL+.5; if(a1 != a2) *((int*)5) = 0; printf("%.0lf\n", arc-dist); } } ]]>