hdu 1598 find the most comfortable road

路径从小到大排序，然后开始枚举边，构建最小生成树，如果包含了要求的起点和终点，即  father【u】==father【v】 ，求得此时的差值，取个最小值就是了

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>#define inf 999999999using namespace std;struct edge{int u,v,w;}e[1100];int father[500000];int n,m;bool cmp(const edge &a,const edge &b){return a.w<b.w;};int findroot(int p){if(father[p]!=p)father[p]=findroot(father[p]);return father[p];}void unionset(int p,int q){father[q]=p;}int main(){while(scanf("%d%d",&n,&m)!=EOF){int i,j;for(i=0;i<m;i++){scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);}sort(e,e+m,cmp);int q;scanf("%d",&q);while(q--){int u,v;int ans=inf;scanf("%d%d",&u,&v);for(i=0;i<m;i++){for(j=0;j<=n;j++)father[j]=j;for(j=i;j<m;j++){int a=findroot(e[j].u);int b=findroot(e[j].v);if(a!=b)father[b]=a;if(findroot(u)==findroot(v)){ans=min(ans,e[j].w-e[i].w);}}}if(ans==inf)printf("-1\n");else printf("%d\n",ans);}}return 0;}