POJ 2263 最短路 路径上最小值

修改floyd

mp[i][j] = MAX(mp[i][j], MIN(mp[i][k], mp[k][j]));

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef  long long LL;
const double PI = acos(-1.0);

template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}
template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}

const int N = 111;
const int M = 11111;
const LL MOD = 1000000007LL;
const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
const int INF = 0x3f3f3f3f;


int mp[222][222];


int main()
{
    int n, m;
    int cas = 1;
    while (cin >> n >> m, n && m)
    {
        string name1, name2;
        int i, j, k, u, v, w, cnt = 1;
        map < string, int> index;
        memset(mp, 0, sizeof(mp));
        for (i = 0; i < m; ++i)
        {
            cin >> name1 >> name2 >> w;
            if (!index[name1]) u = index[name1] = cnt++;
            else u = index[name1];
            if (!index[name2]) v = index[name2] = cnt++;
            else v = index[name2];
            mp[u][v] = mp[v][u] = w;
        }
        for (k = 1; k <= n; ++k)
            for (i = 1; i <= n; ++i)
                for (j = 1; j <= n; ++j)
                {
                    mp[i][j] = MAX(mp[i][j], MIN(mp[i][k], mp[k][j]));
                }
        cin >> name1 >> name2;
        u = index[name1]; v = index[name2];
//        printf("%d  %d\n",u, v);
        printf("Scenario #%d\n%d tons\n\n", cas++, mp[u][v]);
    }
    return 0;
}