LA 3027 合作网络 并查集

题目链接：

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1028

比较简单，用数组d[i]表示结点i和祖先结点的距离，查询时压缩路径并更新d[]数组。

刘汝佳大白书（P193）：

贴代码：

#include<cstdio>
#include<algorithm>
const int N = 20004;
int pa[N],d[N];
int findset(int x)
{
    if(pa[x] == -1) return x;
    else
    {
        int root = findset(pa[x]);
        d[x] += d[pa[x]];
        return pa[x] = root;
    }
}
int main()
{
//    freopen("in.txt","r",stdin);
    int t,n,u,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        char s[4];
        for(int i=0; i<=n; ++i) pa[i]=-1,d[i]=0;
        while(scanf("%s",s),s[0] != 'O')
        {
            if(s[0] == 'E')
            {
                scanf("%d",&u);
                findset(u);
                printf("%d
",d[u]);
            }
            else
            {
                scanf("%d%d",&u,&v);
                pa[u] = v;
                d[u] = abs(u-v)%1000;
            }
        }
    }
    return 0;
}