POJ 2263 最短路 路径上最小值

#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;
}