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; 
}