POJ 1703：Find them, Catch them（带权的并查集）

标签：poj

Description

Input

Output

Sample Input

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

Sample Output

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

题意：

城市里面有2个犯罪团伙，罪犯都属于这两个团伙。现在给你2个罪犯，判断他们是不是一个犯罪团伙。如果当前关系不确定，就输出not sure yet.

并查集的应用，很A bug‘s life基本一样。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>

using namespace std;

const int maxn = 100005;
int cas, n, m;
int parent[maxn], relation[maxn];
char str[2];
int a, b;

int find(int x)
{
    if (x == parent[x])
        return x;
    int px = find(parent[x]);
    relation[x] = ((relation[x] + relation[parent[x]])%2);
    return parent[x] = px;
}

void init()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        parent[i] = i;
        relation[i] = 0;
    }
}

void kruskal(int a, int b)
{
    if(n==2 && str[0]=='A')
//题意是说只有两个人时，应该是属于不同的团伙。。但是数据没有这个。。就算不加也A了、、
        printf("In different gangs.\n");
        else
        {
            int pa = find(a);
            int pb = find(b);
            if (pa != pb)
            {
                if (str[0] == 'D')
                {
                    parent[pa] = pb;
                    if (relation[b] == 0)
                        relation[pa] = 1 - relation[a];//表示不同
                    else
                        relation[pa] = relation[a];//相同
                }
                else printf("Not sure yet.\n");
            }
            else
            {
                if (str[0] == 'A')
                {
                    if (relation[a] != relation[b])
                        printf("In different gangs.\n");
                    else
                        printf("In the same gang.\n");
                }
            }
        }
}

int main()
{
    scanf("%d", &cas);
    while (cas--)
    {
        init();
        while (m--)
        {
            scanf("%s%d%d", str, &a, &b);
            kruskal(a, b);
        }
    }
    return 0;
}

POJ 1703：Find them, Catch them（带权的并查集）

标签：poj

原文地址：http://blog.csdn.net/u013487051/article/details/38031803