标签：poj1703

Find them, Catch them

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4

Not sure yet.
In different gangs.
In the same gang.

 维护关系的并查集。pre[i]表示i为第一个集合,pre[i+N]表示i为第2个集合。由于给定的关系是相对的，所以每次同时维护两种情况。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;

#define maxn 100005

int pre[maxn << 1];

int ufind(int k) {
    int a = k, b;
    while(pre[k]) k = pre[k];
    while(a != k) {
        b = pre[a];
        pre[a] = k;
        a = b;
    }
    return k;
}

bool same(int a, int b) {
    return ufind(a) == ufind(b);
}

bool unite(int a, int b) {
    a = ufind(a);
    b = ufind(b);
    if(a == b) return false;
    pre[a] = b;
    return true;
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    int T, N, M, i, a, b;
    char ch[2];
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &N, &M);
        memset(pre, 0, sizeof(int) * (2 * N + 1));
        while(M--) {
            scanf("%s%d%d", ch, &a, &b);
            if(*ch == 'D') {
                unite(a, b + N);
                unite(a + N, b);
            } else {
                if(same(a, b + N))
                    printf("In different gangs.\n");
                else if(same(a, b)) 
                    printf("In the same gang.\n");
                else printf("Not sure yet.\n");
            }
        }
    }
    return 0;
}

POJ1703 Find them, Catch them 【并查集】

标签：poj1703

原文地址：http://blog.csdn.net/chang_mu/article/details/41675475