HDU 3038 How Many Answers Are Wrong

题解：用加权并查集，将小的节点作为父节点，每一次压缩路径时传递和的信息，如果已有信息存在，判断是否是正确的即可：

#include <cstdio> 

int n, m, data, ans; 

int f[200010],sum[200010]; 

int sf(int x){ 

    int t; 

    if(x==f[x])return f[x]; 

    t=f[x]; 

    f[x]=sf(f[x]);  

    sum[x]+=sum[t];

    return f[x]; 

} 

int Union(int x, int y){ 

    int a, b; 

    a=sf(x); 

    b=sf(y); 

    if(a==b){ 

        if(sum[y]!=sum[x]+data) 

            ans++; 

        return 0; 

    } 

    if(a>b){ 

        sum[a]=sum[y]-sum[x]-data; 

        f[a]=b; 

    } 

    else{ 

        sum[b]=sum[x]+data-sum[y]; 

        f[b]=a; 

    } 

    return 0; 

} 

void init(){ 

    int i; 

    for(i=0;i<=n;i++){ 

        f[i]=i; 

        sum[i]=0; 

    } 

} 

int main(){ 

    int i, j, left, right; 

    while(scanf("%d%d", &n, &m) != EOF){ 

        ans = 0; 

        init(); 

        for(i = 0 ; i < m ; i++){ 

            scanf("%d%d%d",&left,&right,&data); 

            Union(left-1,right); 

        } 

        printf("%d\n", ans); 

    } 

    return 0; 

}