题目链接：kindergarten

最大独立集 = 总的点数 - 最小覆盖集 = 总的点数 - 最大流

匈牙利算法

#include<cstdio>
#include<cstring>
using namespace std;
#define N 205       
int g[N][N];
int match[N];
int vis[N];
int n,m,k;
int dfs(int u)
{
    for(int i=1;i<=m;i++)
        if(!vis[i] && !g[u][i]){
            vis[i]=1;
            if(match[i]==-1 || dfs(match[i])){
                match[i]=u;
                return 1;                
            }
        }
    return 0;
}
int hungary()
{
    int ret=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<=n;i++){
        memset(vis,0,sizeof(vis));
        ret+=dfs(i);
    }
    return ret;
}
int main(void)
{
    int a,b;
    int cas=1;
    while(scanf("%d%d%d",&n,&m,&k)&&(n+m+k))
    {
        memset(g,0,sizeof(g));
        for(int i=0;i<k;i++){
            scanf("%d%d",&a,&b);
            g[a][b]=1;
        }
        printf("Case %d: %d\n",cas++,n+m-hungary());
    }
    return 0;
}

网络流样例没过。。。很难受，先挖个坑吧。。。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 20000;
const int INF = 1e9;
struct Node{
	int ne;
	int to;
	int w;
}e[N<<1];
int head[N],deth[N];
int x,y,val,cnt,n,m,k,st,en;
void init()
{
	memset(head,-1,sizeof(head));
	cnt = 0;
	st = 0;
	en = n + m + 10;
}
void add(int u,int v,int val)
{
	e[cnt].to = v;
	e[cnt].ne = head[u];
	e[cnt].w = val;
	head[u] = cnt ++;
}
bool bfs()
{
	memset(deth,0,sizeof(deth));
	queue<int>q;
	q.push(st);
	deth[st] = 1;
	while(!q.empty())
	{
		int now = q.front();
		q.pop();
		for(int i=head[now];~i;i=e[i].ne)
		{
			int to = e[i].to;
			if(deth[to] == 0 && e[i].w > 0)
			{
				deth[to] = deth[now] + 1;
				q.push(to);
			}
		}
	}
	if(deth[en] == 0)
		return 0;
	return 1;
}
int dfs(int u,int dist)
{
	if(u == en)
	return dist;
	int temp = dist;
	for(int i=head[u];~i;i=e[i].ne)
	{
		int to = e[i].to;
		if(deth[to] == deth[u] + 1 && e[i].w)
		{
			int di = dfs(to,min(e[i].w,temp));
			if(di > 0)
			{
				temp -= di;
				e[i].w -= di;
				e[i^1].w += di;
				if(temp <= 0)
				return dist; 
			}
		}
	}
	return dist - temp;
}
int dinic()
{
	int ans = 0;
	while(bfs())
	{
		while(int di = dfs(st,INF))
			ans += di;
	}
	return ans;
}
int main()
{
	int tot = 0;
	while(scanf("%d%d%d",&n,&m,&k) && n+m+k)
	{
		init();
		for(int i=1;i<=k;i++)
		{
			scanf("%d%d",&x,&y);
			add(x,y,1);
			add(y,x,0);
		}
		for(int i=1;i<=n;i++)
		{
			add(st,i,1);
			add(i,st,0);
		}
		for(int i=1;i<=m;i++)
		{
			add(i+n,en,1);
			add(en,i+n,0);
		}
		int kk = dinic(); 
		printf("Case %d: %d\n",++tot,n+m-kk);
	}
	return 0;
}

还有，这个题竟然卡网络流！！！

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 200;
const int INF = 1e9;
struct Node{
	int ne;
	int to;
	int w;
}e[N<<1];
int head[N],deth[N];
int mp[N][N];
int x,y,val,cnt,n,m,k,st,en;
void init()
{
	memset(head,-1,sizeof(head));
	memset(mp,0,sizeof(mp));
	cnt = 0;
	st = 0;
	en = n + m + 10;
}
void add(int u,int v,int val)
{
	e[cnt].to = v;
	e[cnt].ne = head[u];
	e[cnt].w = val;
	head[u] = cnt ++;
}
bool bfs()
{
	memset(deth,0,sizeof(deth));
	queue<int>q;
	q.push(st);
	deth[st] = 1;
	while(!q.empty())
	{
		int now = q.front();
		q.pop();
		for(int i=head[now];~i;i=e[i].ne)
		{
			int to = e[i].to;
			if(deth[to] == 0 && e[i].w > 0)
			{
				deth[to] = deth[now] + 1;
				q.push(to);
			}
		}
	}
	if(deth[en] == 0)
		return 0;
	return 1;
}
int dfs(int u,int dist)
{
	if(u == en)
	return dist;
	int temp = dist;
	for(int i=head[u];~i;i=e[i].ne)
	{
		int to = e[i].to;
		if(deth[to] == deth[u] + 1 && e[i].w)
		{
			int di = dfs(to,min(e[i].w,temp));
			if(di > 0)
			{
				temp -= di;
				e[i].w -= di;
				e[i^1].w += di;
				if(temp <= 0)
				return dist; 
			}
		}
	}
	return dist - temp;
}
int dinic()
{
	int ans = 0;
	while(bfs())
	{
		while(int di = dfs(st,INF))
			ans += di;
	}
	return ans;
}
int main()
{
	int tot = 0;
	while(scanf("%d%d%d",&n,&m,&k) && n+m+k)
	{
		init();
		for(int i=1;i<=k;i++)
		{
			scanf("%d%d",&x,&y);
			mp[x][y] = 1;
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				if(!mp[i][j])
				{
					add(i,j+n,1);
					add(j+n,i,0);
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			add(st,i,1);
			add(i,st,0);
		}
		for(int i=1;i<=m;i++)
		{
			add(i+n,en,1);
			add(en,i+n,0);
		}
		int kk = dinic(); 
		printf("Case %d: %d\n",++tot,n+m-kk);
	}
	return 0;
}