题解:

二分图匹配

看看是否能达到目标

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=;
int num,ne[N],fi[N],zz[N],k,cnt,a[][],f[N],match[N],x,y,n,m;
void jb(int x,int y)
{
ne[++num]=fi[x];
fi[x]=num;
zz[num]=y;
ne[++num]=fi[y];
fi[y]=num;
zz[num]=x;
}
int dfs(int x)
{
for (int i=fi[x];i;i=ne[i])
if (!f[zz[i]])
{
f[zz[i]]=;
if (!match[zz[i]]||dfs(match[zz[i]]))
{
match[zz[i]]=x;
return ;
}
}
return ;
}
int main()
{
while (~scanf("%d%d%d",&m,&n,&k))
{
memset(fi,,sizeof fi);
memset(a,,sizeof a);
memset(match,,sizeof match);
num=cnt=;
for (int i=;i<=k;i++)
{
scanf("%d%d",&x,&y);
a[y][x]=-;
}
if ((n*m-k)%)
{
puts("NO");
continue;
}
for (int i=;i<=m;i++)
for (int j=;j<=n;j++)
if (!a[i][j])a[i][j]=++cnt;
for (int i=;i<=m;i++)
for (int j=;j<=n;j++)
if (a[i][j]!=-)
{
if (a[i+][j]!=-&&i!=m)jb(a[i][j],a[i+][j]);
if (a[i][j+]!=-&&j!=n)jb(a[i][j],a[i][j+]);
}
int ans=;
for (int i=;i<=cnt;i++)
{
memset(f,,sizeof f);
ans+=dfs(i);
}
ans/=;
if (ans*==cnt)puts("YES");
else puts("NO");
}
}
05-26 02:45