pku 1125 Stockbroker Grapevine 第一周训练——最短路

http://poj.org/problem?id=1125

刚开始以为是求最小生成树，可是打完代码一看结果根本不对。最后看了一下discuss原来是的那个一个点和m个点连接时，可以同时传送信息。所以只要转换成dijkstra求原点到所有点的最短然后找出最大时间，就是从原点到最后一个点的时间长度了，，循环遍历所有点当做原点情况，然后找出最佳值。。

View Code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define maxn 107
using namespace std;
int dis[maxn],map[maxn][maxn];
bool vt[maxn];
const int inf = 9999999;
int ans,tmp;
int n;
void init()
{
    int i,j;
    for (i = 0; i <= n; ++i)
    {
        for (j = 0; j <= n; ++j)
        {
            if (i == j) map[i][j] = 0;
            else map[i][j] = inf;
        }
    }
}
int Dijkstra(int s)
{
    int i,j,k,min;
    int sum = 0;
    for (i = 1; i <= n; ++i)
    {
        dis[i] = map[s][i];
        vt[i] = false;
    }
    vt[s] = true;
    for (k = 1; k < n; ++k)
    {
        j = s; min = inf;
        for (i = 1; i <= n; ++i)
        {
            if (!vt[i] && dis[i] < min)
            {
                min = dis[i];
                j = i;
            }
        }
        vt[j] = true;
        for (i = 1; i <= n; ++i)
        {
            if (!vt[i] && dis[i] > dis[j] + map[j][i])
            dis[i] = dis[j] + map[j][i];
        }
    }
    for (i = 1; i <= n; ++i)//循环找出最长的时间段
    {
        if (sum < dis[i])
        sum = dis[i];
    }
    if (sum == inf)
    return -1;
    else
    return sum;
}
int main()
{
    int i,j,m,a,b;
    while (~scanf("%d",&n))
    {
        if (!n) break;
        init();
        for (i = 1; i <= n; ++i)
        {
            cin>>m;
            for (j = 0; j < m; ++j)
            {
                cin>>a>>b;
                map[i][a] = b;
            }
        }
        int ct = 0,t = 0;
        ans = inf;
        for (i = 1; i <= n; ++i)//枚举所有点为原点
        {
            int tmp = Dijkstra(i);
            if (tmp == -1)//判断是否可达
            {
                ct++;
            }
            else if (tmp < ans)//记录最佳值
            {
                ans = tmp;
                t = i;
            }
        }
        if (ct == n)
        {
            printf("disjoint\n");
        }
        else
        printf("%d %d\n",t,ans);
    }
    return 0;
}