poj2492 带权并查集 三种解法

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题意：给定n个昆虫和m个判断x, y，表示x和y是异性。求判断是否错误。

法一：带权并查集

#include <cstdio>

using namespace std;
const int N = 2010;
int fa[N],relation[N];//0:相同，1不同
int n, m;

int findf(int x)
{
    if(x != fa[x])
    {
        int f = findf(fa[x]);
        relation[x] = (relation[x]+relation[fa[x]])%2;//与朴素的不同
        fa[x] = f;
    }
    return fa[x];
}

bool merge_(int x, int y)
{
    int f1 = findf(x);
    int f2 = findf(y);
    if(f1==f2)
    {
        if((relation[x]+relation[y])%2==0)
            return 1;
        return 0;
    }
    fa[f2] = f1;
    relation[f2] = (relation[x]+relation[y]+1)%2;//保证更新后x,y为异性
    return 0;
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= n; ++i)
        {
            fa[i] = i;
            relation[i] = 0;
        }
        bool flag = 0;
        while(m--)
        {
            int x, y;
            scanf("%d %d", &x, &y);
            if(merge_(x, y))
                flag = 1;
        }
        if(flag) printf("Scenario #%d:\nSuspicious bugs found!\n\n", cas);
        else printf("Scenario #%d:\nNo suspicious bugs found!\n\n", cas);
    }
    return 0;
}

法二：把一个点i分成两个点i和i+n，一个表示相同，一个表示不同

#include <cstdio>

using namespace std;
const int N = 2010;
int fa[N<<1];
int n, m;

int findf(int x)
{
    return x==fa[x]?x:fa[x]=findf(fa[x]);
}
void merge_(int x, int y)
{
    int f1 = findf(x);
    int f2 = findf(y);
    if(f1!=f2)
        fa[f1] = f2;
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= 2*n; ++i) fa[i] = i;
        bool flag = 0;
        while(m--)
        {
            int x, y;
            scanf("%d %d", &x, &y);
            if(findf(x)==findf(y)) flag = 1;
            else
            {
                merge_(x, y+n);
                merge_(x+n, y);
            }
        }
        if(flag) printf("Scenario #%d:\nSuspicious bugs found!\n\n", cas);
        else printf("Scenario #%d:\nNo suspicious bugs found!\n\n", cas);
    }
    return 0;
}

法三：dif[x]表示与x性别不同的昆虫

#include <cstdio>

using namespace std;
const int N = 2010;
int fa[N<<1],dif[N<<1];
int n, m;

int findf(int x)
{
    return x==fa[x]?x:fa[x]=findf(fa[x]);
}

void merge_(int x, int y)
{
    int f1 = findf(x);
    int f2 = findf(y);
    if(f1!=f2)
        fa[f1] = f2;
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= 2*n; ++i)
        {
            fa[i] = i;
            dif[i] = 0;
        }
        bool flag = 0;
        while(m--)
        {
            int x, y;
            scanf("%d %d", &x, &y);
            if(findf(x)==findf(y)) flag = 1;
            else if(dif[x]+dif[y]==0)
            {
                dif[x] = y;
                dif[y] = x;
            }
            else if(!dif[x])
            {
                dif[x] = y;
                merge_(x, dif[y]);
            }
            else if(!dif[y])
            {
                dif[y] = x;
                merge_(y, dif[x]);
            }
            else
            {
                merge_(x, dif[y]);
                merge_(y, dif[x]);
            }
        }
        if(flag) printf("Scenario #%d:\nSuspicious bugs found!\n\n", cas);
        else printf("Scenario #%d:\nNo suspicious bugs found!\n\n", cas);
    }
    return 0;
}