紫书 例题 11-5 UVa 10048 (Floyd求最大权值最小的路径)

这道题是Floyd的变形


改成d[i][j] = min(d[i][j], max(d[i][k], d[k][j]))就好了。

#include<cstdio>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 112;
int d[MAXN][MAXN], n, m, q;

int main()
{
	int kase = 0;
	while(~scanf("%d%d%d", &n, &m, &q) && n)
	{
		REP(i, 0, n)
			REP(j, 0, n)
				d[i][j] = (i == j ? 0 : 1e9);
			
		while(m--)
		{
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			u--; v--;
			d[v][u] = d[u][v] = min(d[u][v], w);
		}
		
		REP(k, 0, n)
			REP(i, 0, n)
				REP(j, 0, n)
					d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));
		
		if(kase) puts("");
		printf("Case #%d\n", ++kase);
		while(q--)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			u--; v--;
			if(d[u][v] == 1e9) puts("no path");
			else printf("%d\n", d[u][v]);
		}  
	}
	
	return 0;	
} 

posted @ 2018-05-25 19:33  Sugewud  阅读(133)  评论(0编辑  收藏  举报