【poj1962】 Corporative Network

http://poj.org/problem?id=1962 (题目链接)

时隔多年又一次写带权并查集。

题意

　　n个节点，若干次询问，I x y表示从x连一条边到y，权值为｜x-y｜%1000；E x表示询问x到x所指向的终点的距离。

Solution

　　很裸的带权并查集。

代码

// poj1962
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#define LL long long
#define MOD 1000
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
int getint() {
    int f=1,x=0;char ch=getchar();
    while (ch<='0' || ch>'9') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=20010;
int r[maxn],fa[maxn],n;

int find(int x) {
    if (x==fa[x]) return x;
    int f=find(fa[x]);
    r[x]=r[x]+r[fa[x]];
    fa[x]=f;
    return f;
}
int main() {
    int T;scanf("%d",&T);
    while (T--) {
        scanf("%d",&n);
        for (int i=1;i<=n;i++) fa[i]=i,r[i]=0;
        char ch[10];
        while (scanf("%s",ch)!=EOF) {
            int x,y;
            if (ch[0]=='O') break;
            else if (ch[0]=='E') {
                scanf("%d",&x);
                find(x);
                printf("%d
",r[x]);
            }
            else {
                scanf("%d%d",&x,&y);
                fa[x]=y;
                r[x]=abs(x-y)%MOD;
            }
        }
    }
    return 0;
}