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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 19226 | Accepted: 8775 |
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
Output
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
Hint
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int map[1001][1001],w[1001][1001]; int dis[1001],Team[5000]; bool exist[1001]; int n,m,t; // exist 为 true 则不可进队 void SPFA(int s) { int head=0,tail=1,k=0; Team[head]=s,exist[s]=true;dis[s]=0; while(head<tail) { k=Team[head]; exist[k]=false; for(int i=1;i<=n;i++) { if((map[k][i]>0)&&(dis[i]>dis[k]+map[k][i])) { dis[i]=map[k][i]+dis[k]; if(exist[i]==false) { Team[tail++]=i; exist[i]=true; } } } head++; } } int main() { cin>>n>>m>>t; memset(map,0x3f,sizeof map ); memset(w,0x3f,sizeof w ); memset(dis,0x3f,sizeof dis ); for(int i=1;i<=m;i++) { int x,y,z; cin>>x>>y>>z; map[x][y]=z; w[x][y]=z; } SPFA(t);// 从起点开始到其他每个点找一遍最短路 for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(w[i][j]>w[i][k]+w[k][j]) w[i][j]=w[i][k]+w[k][j]; } } } int max; for(int i=1;i<=n;i++) { if((i==1)||(dis[i]+w[i][t]>max)) max=dis[i]+w[i][t]; } printf("%d",max); return 0; }
思路：一遍SPFA求出起点到其他所有点的最短路，一遍Floyed求出所有点到起点的最短路，两者相加，和最大者即为anwser...
BZOJ 1631==USACO 2007== POJ 3268 Cow Party奶牛派对
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原文地址：http://www.cnblogs.com/zhishenduchuang/p/5877983.html
