HDU 2444 The Accomodation of Students(判断是否是二分图)

题目链接

题意:n个学生,m对关系,每一对互相认识的能住一个房间。问否把这些学生分成两组,要求每组的学生都互不认识。求最多需要多少个房间。

是否能分成两组?也就是说判断是不是二分图,判断二分图的办法,用染色法

把初始点染成黑色,然后与之相连的染成白色,重复,使路径黑白相间,

如果当前点的颜色和与他相连点的颜色相同时,则说明这个图不是二分图

求最多需要多少个房间?也就是求最大匹配数。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
#define init(a) memset(a,0,sizeof(a))
#define PI acos(-1,0)
using namespace std;
const int maxn = 310;
const int maxm = 100001;
#define lson left, m, id<<1
#define rson m+1, right, id<<1|1
#define min(a,b) (a>b)?b:a
#define max(a,b) (a>b)?a:b
const int N = 50010;
int ma[maxn][maxn];
int line[maxn],color[maxn];
bool vis[maxn];
int k,n,m;
bool flag;
int DFS(int u)
{
    for(int  v = 1;v<=n;v++)
    {
        if(ma[u][v]==1 && !vis[v])
        {
            vis[v] = 1;
            if(line[v]==-1 || DFS(line[v]))
            {
                line[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
void dfs(int u,int st)
{
    if(flag == false)
        return ;
    for(int v = 1;v <= n;v++)
    {
        if(ma[u][v])
    {
        if(!color[v])
        {
            color[v]= -st;//黑染白 / 白染黑
            dfs(v,color[v]);
        }
        else if(color[v]==st)
        {
            flag = false;
            return;
        }

    }
    }

}
bool judge()
{
    flag = true;
    color[1] = 1 ;//1代表黑色,-1代表白色
    dfs(1,1) ;
    return flag;
}
int K_M()
{
    int ans = 0;
    memset(line,-1,sizeof(line));
    for(int i  = 1;i<=n;i++)
    {
        init(vis);
        ans += DFS(i);
    }
    return ans;
}
int main()
{
    int a,b;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init(ma);
        memset(color,0,sizeof(color));
        for(int i = 1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            ma[a][b]  = 1;
        }
        if(!judge())
        {
            puts("No");
            continue;
        }
        int ans = K_M();
        printf("%d\n",ans);
    }
    return 0;
}

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时间: 2024-10-13 03:12:26

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