upc 5220 A Possible Tree

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带权并查集

代码如下

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int t,n,c,x,y;
long long val,r[100005];
int p[100005];
int f(int x)
{
    if(x==p[x]) return x;
    int fa=p[x];
    p[x]=f(p[x]);
    r[x]^=r[fa];
    return p[x];
}
int main()
{
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n,&c);
        for(int i=0;i<n-1;i++) {
            scanf("%d%d",&x,&y);
        }
        for(int i=1;i<=n;i++) {
            r[i]=0; p[i]=i;
        }
        int ans=1e9;
        for(int i=0;i<c;i++) {
            scanf("%d%d%lld",&x,&y,&val);
            if(ans!=1e9) continue;
            int fx=f(x),fy=f(y);
            if(fx==fy&&(r[x]^r[y])!=val) {
                ans=i;
            }
            else{
                p[fx]=fy;
                r[fx]=r[x]^r[y]^val;
            }
        }
        printf("%d\n",ans);
    }
}

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转载自blog.csdn.net/qq_36651153/article/details/82108618
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