POJ 3268 Silver Cow Party
阿新 • • 發佈：2018-08-21
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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 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 XLines 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 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 |
從幾個點 到x 再從x回來
這個有個小技巧 ，按照一般寫會超時 各個點到一個點x的最短距離 可以換成x到各個點的，只需把矩陣反轉一下再跑一遍圖論算法就可以了
#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; }
POJ 3268 Silver Cow Party