Slim Span POJ

最小生成树最大边和最小边之差最小

题意：
求最小生成树最大边和最小边的最小差值

思路
用kruskal 算法，枚举最小边即可

代码

/* * Author       :  Echo * Email        :  [e-mail]   * Description  :    * Created Time :  2017/10/10 19:25:26 * Last Modify  :  2017/10/14 16:27:10 * File Name    :  \yanga11ang\write.cpp */#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <queue>using namespace std;const int maxn=110;const int maxm=1e4+100;const double INF=1e9;struct node{    int u,v,w;}an[maxm];int n,m;int fa[maxn];int findp(int a){    if(fa[a]==a) return a;    fa[a]=findp(fa[a]);    return fa[a];}int kruskal(int star){    for(int i=1;i<=n;i++){        fa[i]=i;    }    int cnt=0;    int maxx=-1e9,minn=1e9;    for(int i=star;i<=m;i++){        int u=an[i].u;        int v=an[i].v;        int up=findp(u);        int vp=findp(v);        if(up==vp) continue;        fa[up]=vp;        cnt++;        if(maxx<an[i].w) maxx=an[i].w;        if(minn>an[i].w) minn=an[i].w;    }    if(cnt!=n-1) return-1;    return maxx-minn;}bool cmp(node a,node b){    return a.w<b.w;}int main(){    while(~scanf("%d%d",&n,&m)){        if(n==m&&n==0)break;        for(int i=1;i<=m;i++){            scanf("%d%d%d",&an[i].u,&an[i].v,&an[i].w);        }        sort(&an[1],&an[1]+m,cmp);        int ans=1e9;        for(int i=1;i<=m;i++){            int res=kruskal(i);            if(res==-1) break;            if(ans>res) ans=res;        }        if(ans==1e9) printf("-1\n");        else printf("%d\n",ans);    }    return 0;}/* */