Silver Cow Party
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 13982 | Accepted: 6307 |
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1: One integer: the maximum of time any one cow must walk.
Sample Input
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
Sample Output
10
因为迪杰斯特拉求得是一点到其他所有点的最短路径，所以先把起点设成X,跑一次之后算出了每头牛回去的最短路径，然后把边反转，入边变成出边，出边变成入边，然后再把起点设成X，跑一次之后就算出了每头牛来的最短路径，最后把每头牛来回的最短路径加起来，取所有牛的最大值。写得太挫，明天补发其他解法。
#include <iostream> #include <cstdio> #include <queue> #include <vector> using namespace std; const int SIZE = 1005; const int INF = 0x6fffffff; int MAP[SIZE][SIZE]; bool VIS[SIZE][SIZE]; int BOX[SIZE]; int D[SIZE],S[SIZE]; int N,M,X; struct Comp { bool operator ()(int & a,int & b) { return D[a] > D[b]; } }; struct Node { int vec,cost; }; vector<Node> G[SIZE]; void dijkstra(int); int main(void) { Node temp; int from; while(scanf("%d%d%d",&N,&M,&X) != EOF) { fill(&MAP[0][0],&MAP[SIZE - 1][SIZE - 1],INF); fill(&VIS[0][0],&VIS[SIZE - 1][SIZE - 1],false); fill(BOX,BOX + SIZE - 1,0); for(int i = 1;i <= N;i ++) G[i].clear(); for(int i = 0;i < M;i ++) { scanf("%d%d%d",&from,&temp.vec,&temp.cost); G[from].push_back(temp); MAP[from][temp.vec] = temp.cost; } dijkstra(X); for(int i = 1;i <= N;i ++) BOX[i] = D[i]; for(int i = 1;i <= N;i ++) G[i].clear(); for(int i = 1;i <= N;i ++) for(int j = 1;j <= N;j ++) { temp.vec = j; temp.cost = MAP[j][i]; G[i].push_back(temp); } dijkstra(X); int ans = -1; for(int i = 1;i <= N;i ++) { BOX[i] += D[i]; ans = ans > BOX[i] ? ans : BOX[i]; } printf("%d ",ans); } return 0; } void dijkstra(int s) { fill(D,D + SIZE,INF); fill(S,S + SIZE,false); priority_queue<int,vector<int>,Comp> que; D[s] = 0; que.push(s); while(!que.empty()) { int cur = que.top(); que.pop(); S[cur] = true; for(int i = 0;i < G[cur].size();i ++) if(!S[G[cur][i].vec] && D[G[cur][i].vec] > D[cur] + G[cur][i].cost) { D[G[cur][i].vec] = D[cur] + G[cur][i].cost; que.push(G[cur][i].vec); } } }
补发贝尔曼和SPFA
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #include <vector> 5 using namespace std; 6 7 const int INF = 0x6fffffff; 8 const int SIZE = 1005; 9 int N,M,X; 10 int MAP[SIZE][SIZE],D[SIZE],BOX[SIZE]; 11 bool VIS[SIZE]; 12 struct Node 13 { 14 int vec,cost; 15 }; 16 vector<Node> G[SIZE]; 17 18 void SPFA(int); 19 bool relax(int,int,int); 20 int main(void) 21 { 22 int from,to,cost,ans; 23 Node temp; 24 25 while(scanf("%d%d%d",&N,&M,&X) != EOF) 26 { 27 fill(BOX,BOX + SIZE,0); 28 fill(&MAP[0][0],&MAP[SIZE - 1][SIZE - 1],INF); 29 30 for(int i = 0;i < M;i ++) 31 { 32 scanf("%d%d%d",&from,&to,&cost); 33 MAP[from][to] = cost; 34 temp.vec = to; 35 temp.cost = cost; 36 G[from].push_back(temp); 37 } 38 SPFA(X); 39 for(int i = 1;i <= N;i ++) 40 BOX[i] += D[i]; 41 for(int i = 1;i <= N;i ++) 42 { 43 G[i].clear(); 44 for(int j = 1;j <= N;j ++) 45 { 46 if(MAP[j][i] == INF) 47 continue; 48 temp.vec = j; 49 temp.cost = MAP[j][i]; 50 G[i].push_back(temp); 51 } 52 } 53 SPFA(X); 54 ans = -1; 55 for(int i = 1;i <= N;i ++) 56 { 57 BOX[i] += D[i]; 58 ans = ans > BOX[i] ? ans : BOX[i]; 59 } 60 printf("%d ",ans); 61 } 62 63 return 0; 64 } 65 66 void SPFA(int s) 67 { 68 fill(D,D + SIZE,INF); 69 fill(VIS,VIS + SIZE,false); 70 D[s] = 0; 71 queue<int> que; 72 que.push(s); 73 VIS[s] = true; 74 75 while(!que.empty()) 76 { 77 int cur = que.front(); 78 que.pop(); 79 VIS[cur] = false; 80 81 for(int i = 0;i < G[cur].size();i ++) 82 { 83 if(relax(cur,G[cur][i].vec,G[cur][i].cost)) 84 { 85 que.push(G[cur][i].vec); 86 VIS[G[cur][i].vec] = true; 87 } 88 } 89 } 90 } 91 92 bool relax(int from,int to,int cost) 93 { 94 if(D[to] > D[from] + cost) 95 { 96 D[to] = D[from] + cost; 97 return true; 98 } 99 return false; 100 }
Bellman_ford
#include <iostream> #include <cstdio> using namespace std; const int INF = 0x6fffffff; const int SIZE = 1005; int D[SIZE],BOX[SIZE];; int N,M,X; struct Node { int from,to,cost; }G[SIZE * 100]; void Bellman_ford(int); bool relax(int,int,int); int main(void) { int ans; while(scanf("%d%d%d",&N,&M,&X) != EOF) { fill(BOX,BOX + SIZE,0); for(int i = 0;i < M;i ++) scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost); Bellman_ford(X); for(int i = 1;i <= N;i ++) BOX[i] += D[i]; for(int i = 0;i < M;i ++) swap(G[i].from,G[i].to); Bellman_ford(X); ans = -1; for(int i = 1;i <= N;i ++) { BOX[i] += D[i]; ans = ans > BOX[i] ? ans : BOX[i]; } printf("%d ",ans); } return 0; } void Bellman_ford(int x) { fill(D,D + SIZE,INF); D[x] = 0; bool flag; for(int i = 0;i < N - 1;i ++) { flag = false; for(int j = 0;j < M;j ++) if(relax(G[j].from,G[j].to,G[j].cost)) flag = true; if(!flag) break; } } bool relax(int from,int to,int cost) { if(D[to] > D[from] + cost) { D[to] = D[from] + cost; return true; } return false; }