标签：put order 矩阵 str long swa mes art each
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Silver Cow Party
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. 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 |
从几个点 到x 再从x回来
这个有个小技巧 ，按照一般写会超时 各个点到一个点x的最短距离 可以换成x到各个点的，只需把矩阵反转一下再跑一遍图论算法就可以了
#include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> #define inf 0x3f3f3f using namespace std; int map[1010][1010]; int dis[1010]; int dis1[1010]; int vis[1010]; int n,m,x; void djst(int s) { memset(vis,0,sizeof(vis)); memset(dis,inf,sizeof(dis)); dis[s]=0; while(1) { int k=-1,minn=inf; for(int i=1;i<=n;i++) { if(!vis[i]&&dis[i]<minn) k=i,minn=dis[i]; } if(k==-1) break; vis[k]=1; for(int i=1;i<=n;i++) dis[i]=min(dis[i],dis[k]+map[k][i]); } } int main() { while(scanf("%d %d %d",&n,&m,&x)!=EOF) { memset(map,inf,sizeof(map)); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i==j) map[i][j]=0; else map[i][j]=inf; } } for(int i=0;i<m;i++) { int a,b,c; cin>>a>>b>>c; if(map[a][b]>c) map[a][b]=c; } djst(x); for(int i=1;i<=n;i++) dis1[i]=dis[i]; /* for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { cout<<map[i][j]<<" "; } cout<<endl; } cout<<endl; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { cout<<map[i][j]<<" "; } cout<<endl; } */ for(int i=1;i<=n;i++) { for(int j=1;j<=i;j++) { swap(map[i][j],map[j][i]); } } int ans=0; djst(x); for(int i=1;i<=n;i++) { // cout<<dis[i]<<" "<<dis1[i]<<endl; ans=max(ans,dis[i]+dis1[i]); } cout<<ans<<endl; } return 0; }
标签：put order 矩阵 str long swa mes art each
原文地址：https://www.cnblogs.com/wpbing/p/9508757.html
