uva10048

题目描述：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=22156

/*
solution:   此题可以直接套上floyd算法的模板，但是要把加法改成max
                对于任意一条至少包含两条边的路径，i->j，一定存在k使得i->j
                噪音的最高级等于max(d[i][k], d[k][j]),但i->j路径可能并不唯一，
                所以还要取一条最小的：d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));
author:    LinVan
time:       2016/4/21
*/
#include 
#include 
#include 
#include 

using namespace std;
const int maxC = 100 + 5;
const int maxS = 1000 + 5;
const int maxQ = 10000 + 5;
const int INF = 99999;

int c, s, q, d[maxC][maxC];

void floyd() {
    for(int k = 1; k <= c; k++)
        for(int i = 1; i <= c; i++)
            for(int j = 1; j <= c; j++)
                //if(d[i][k] < INF && d[k][j] < INF)
                    d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int kase = 0;
    while(scanf("%d%d%d", &c, &s, &q) == 3 && c) {
        for(int i = 1; i <= c; i++)
        for(int j = 1; j <= c; j++) {
            if(i == j)      d[i][j] = 0;
            else        d[i][j] = INF;  //初始化
        }

        int a, b, val;
        for(int i = 1; i <= s; i++) {
            scanf("%d%d%d", &a, &b, &val);
            d[a][b] = d[b][a] = val;    //注意数据输入是无向边
        }

        if(kase)    printf("\n");
        printf("Case #%d\n", ++kase);
        floyd();    //floyd主算法

        int u, v;   //查询
        for(int i = 0; i < q; i++) {
            scanf("%d%d", &u, &v);
            if(d[u][v] == INF)      printf("no path\n");
            else        printf("%d\n", d[u][v]);
        }
    }
    return 0;
}