UVa 10048 Audiophobia

题目链接：
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=989
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

// d[i][j]代表从第i点到第j点需要忍受的最少分贝级别
// d[i][j] = -1 代表从第i点到第j点没有路
int d[110][110];

int c, s, q;

int main()
{
int count = 1;
while(scanf("%d %d %d", &c, &s, &q) == 3 && !(c == 0 && s == 0 && q == 0))
{
// 初始化
memset(d, -1, sizeof(d));
for(int i = 1; i <= c; i++)
d[i][i] = 0;

// 读入图
for(int i = 1; i <= s; i++)
{
int x, y, w;
scanf("%d%d%d", &x, &y, &w);
d[x][y] = w;
d[y][x] = w;
}

// 用Floyd-Warshall算法计算结果
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] != -1 && d[k][j] != -1)
{
if(d[i][j] == -1)
d[i][j] = max(d[i][k], d[k][j]);
else
d[i][j] = min(d[i][j], max(d[i][k],d[k][j]));
}
}

if(count > 1)
printf("\n");
printf("Case #%d\n", count);
// 读入查询
for(int i = 1; i <= q; i++)
{
int x, y;
scanf("%d %d", &x, &y);
if(d[x][y] == -1)
printf("no path\n");
else
printf("%d\n", d[x][y]);
}
count++;
}
return 0;
}

本题仍然使用的Floyd-Warshall算法框架来求解d[i][j]为从第i个点至第j个点至少需要忍受的分贝级数。
对于中间的任何一点k来说，i到j的路径如果经过k, 那么d[i][j] = max(d[i][k], d[k][j]).