POJ 1125 Stockbroker Grapevine //Floyd算法

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题目描述

POJ1125 Stockbroker Grapevine

解题思路

找出一个人,使谣言从其传出到所有人用时最短.
(1):用Floyd先算出某人到每个人的最短时间,再对时间取最大 (保证每个人都能收到谣言…) = =|
(2):对每个人执行(1),取时间最小,即为所求.

参考代码

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#define inf 0x3f3f3f3f#define MAX_V 110using namespace std;int f[MAX_V][MAX_V],a[MAX_V];int main(){    int n,num,u,cost;    while(~scanf("%d",&n),n){        memset(f,inf,sizeof(f));        memset(a,0,sizeof(a));        int num;        for(int i = 1;i <= n;i++){            scanf("%d",&num);            while(num--){                scanf("%d %d",&u,&cost);                f[i][u]=cost;            }        }        for(int k = 1;k <= n;k++)            for(int i = 1;i <= n;i++)                for(int j = 1;j <= n;j++)                    if(i!=k && i!=j && j!=k)                        f[i][j] = min(f[i][j],f[i][k]+f[k][j]);        int ans = inf,t;        for(int i = 1;i <= n;i++){            for(int j = 1;j <= n;j++)                if(i != j && (f[i][j]>a[i]))                    a[i] = f[i][j];            if(a[i] < ans){                ans = a[i];                t = i;            }        }        if(ans == inf)printf("disjoint\n");        else printf("%d %d\n",t,ans);    }    return 0;}

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