UVALive - 3027 Corporative Network 并查集+向量

题目大意:有n个节点,初始时每个节点都是独立的,现在给出两种操作

I u v,将节点u的父节点设置为v,u和v两点的距离为abs(u-v) % 1000

E u,询问u节点到根节点的距离

解题思路;按并查集的思路来,不过这题要算上距离,在压缩路径的同时把距离给计算出来

[code]#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define maxn 20010
int p[maxn], dis[maxn], N; 

void init() {
    for(int i = 0; i <= N; i++)
        p[i] = i, dis[i] = 0;
}

int find(int x) {
    if(p[x] != x) {
        int root = find(p[x]);
        dis[x] += dis[p[x]];
        return p[x] = root;
    }
    else 
        return x;
}

int main() {
    int test, x, y;
    char str[10];
    scanf("%d", &test);
    while(test--) {
        scanf("%d", &N);
        init();
        while(1) {
            scanf("%s", str);
            if(str[0] == 'I') {
                scanf("%d%d", &x,&y);
                p[x] = y;
                dis[x] = abs(x - y) % 1000;
            }
            else if(str[0] == 'E') {
                scanf("%d", &x);
                find(x);
                printf("%d\n",dis[x]);  
            }
            else 
                break;
        }
    }
    return 0;
}