poj 1125（最短路）

题意：求从谁发布谣言能使最后得到谣言的人的时间最少。

 

思路：枚举源点求最短路。

 

参考代码：

#include <iostream>#include <cstdio>using namespace std;int edge[105][105],dist[105],n;bool vis[105];#define MAX 10000000int dij(int s){int i,j,k;memset(vis,0,sizeof(vis));for(i=1;i<=n;i++){dist[i]=edge[s][i];}vis[s]=true;for(j=1;j<=n-2;j++){k=1;while(vis[k])k++;int beg=k;for(i=1;i<=n;i++){if(!vis[i]&&dist[i]<dist[k]){k=i;}}vis[k]=true;for(i=beg;i<=n;i++){if(!vis[i]&&dist[k]+edge[k][i]<dist[i]){dist[i]=dist[k]+edge[k][i];}}}k=0;for(i=1;i<=n;i++){if(i!=s&&dist[i]>k){k=dist[i];}}return k;}void solve(){int i,j,k=MAX,ans;for(i=1;i<=n;i++){j=dij(i);if(j<k){k=j;ans=i;}}if(k>=MAX){printf("disjoint/n");}else{printf("%d %d/n",ans,k);}}int main(){//freopen("in.txt","r",stdin);int i,j,k,x;while(scanf("%d",&n)!=EOF&&n){for(i=1;i<=n;i++){for(j=1;j<=n;j++){edge[i][j]=MAX;}}for(i=1;i<=n;i++){scanf("%d",&x);while(x--){scanf("%d%d",&j,&k);edge[i][j]=k;}}solve();}return 0;}