UVa 10048 Audiophobia

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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]);elsed[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");elseprintf("%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]).

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