POJ 1125 Stockbroker Grapevine

 

每次枚举一个顶点，然后计算它到各个点的最大值，然后去顶点中最小的一个。

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <map>
using namespace std;

const int MAXN = 110;
const int INF = 0x3f3f3f3f;

int d[MAXN][MAXN];

int n, m;

void init()
{
    for(int i = 1; i < MAXN; i++)
    {
        for(int j = 1; j < MAXN; j++)
        {
            d[i][j] = (i == j) ? 0:INF;
        }
    }
}

void Floyd()
{
    for(int k = 1; k <= n; k++)
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int read_case()
{
    init();
    scanf("%d", &n);
    if(!n) return 0;
    for(int u = 1; u <= n; u++)
    {
        scanf("%d", &m);
        while(m--)
        {
            int v, w;
            scanf("%d%d", &v, &w);
            d[u][v] = w;
        }
    }
    return 1;
}

void solve()
{
    Floyd();
    int mintime = INF, index = 0;
    for(int i = 1; i <= n; i++)
    {
        int m = -1;
        for(int j = 1; j <= n; j++) if(i != j)
        {
            m = max(m, d[i][j]);
        }
        if(mintime > m)
        {
            mintime = m;
            index = i;
        }
    }
    if(mintime >= INF) printf("disjoint\n");
    else printf("%d %d\n", index, mintime);
}

int main()
{
    while(read_case())
    {
        solve();
    }
    return 0;
}

 

CODE：