POJ 2492 A Bug's Life

并查集确定昆虫的关系，题意有点BT...用一个数组rel代表与其父节点的关系，为0则代表相同，为1代

表不同。如果两者不在一个集合中则合并，在一个集合中则判断种类是否相同，相同则存在同性恋的虫子。

/*Accepted    164K    766MS    C++    1224B    2012-08-24 09:28:15*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

const int MAXN = 1 << 11;
int p[MAXN], rel[MAXN];

int find_set(int x)
{
    int t;
    if(x != p[x])
    {
        t = p[x];
        p[x] = find_set(p[x]);
        rel[x] = (rel[x] + rel[t]) & 1;
    }
    return p[x];
}

int main()
{
    int cas, T;
    int x, y, n, m, i;
    bool flag;
    scanf("%d", &T);
    for(cas = 1; cas <= T; cas ++)
    {
        scanf("%d%d", &n, &m);
        for(i = 1; i <= n; i ++)
        {
            p[i] = i, rel[i] = 0;
        }
        flag = false;
        while(m --)
        {
            scanf("%d%d", &x, &y);
            int nx, ny;
            nx = find_set(x), ny = find_set(y);
            if(nx != ny)
            {
                p[ny] = nx;
                rel[ny] = (rel[y] + rel[x] + 1) & 1;
            }
            else
            {
                if(rel[x] == rel[y])
                    flag = true;
            }
        }
        printf("Scenario #%d:\n", cas);
        if(flag)
            printf("Suspicious bugs found!\n");
        else
            printf("No suspicious bugs found!\n");
        printf("\n");
    }
    return 0;
}