[二分圖最大獨立集] bzoj3175: [Tjoi2013]攻擊裝置

bzoj 3175: [Tjoi2013]攻擊裝置
同 bzoj4808 馬 http://blog.csdn.net/qq_36038511/article/details/79601632

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
    int x,y,next;
}a[410000];
int last[41000],len;
int match[81000],v[81000],c[210][210];
int t;
void build(int x,int y)
{
    len++;
    a[len].x=x;a[len].y=y;a[len].next=last[x];last[x]=len;
}
bool find(int x)
{
    for (int k=last[x];k;k=a[k].next)
    {
        int y=a[k].y;
        if (v[y]!=t)
        {
            v[y]=t;
            if (match[y]==0||find(match[y])==1) 
            {
                match[y]=x;
                return 1;
            }
        }
    }
    return 0;
}
char ch[210];
int main()
{
    int n,sum=0;
    scanf("%d",&n);
    memset(c,-1,sizeof(c));
    for (int i=1;i<=n;i++)
    {
        scanf("%s",ch+1);
        for (int j=1;j<=n;j++) 
        {
            c[i][j]=ch[j]-'0';
            if (c[i][j]==0) sum++;
        }
    }
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
        {
            if (c[i][j]==0)
            {
                int x=(i-1)*n+j;
                if (i-1>=1&&j-2>=1&&c[i-1][j-2]==0) build(x,(i-2)*n+j-2+sum);
                if (i+1<=n&&j-2>=1&&c[i+1][j-2]==0) build(x,i*n+j-2+sum);
                if (i-1>=1&&j+2<=n&&c[i-1][j+2]==0) build(x,(i-2)*n+j+2+sum);
                if (i+1<=n&&j+2<=n&&c[i+1][j+2]==0) build(x,i*n+j+2+sum);
                if (i-2>=1&&j-1>=1&&c[i-2][j-1]==0) build(x,(i-3)*n+j-1+sum);
                if (i-2>=1&&j+1<=n&&c[i-2][j+1]==0) build(x,(i-3)*n+j+1+sum);
                if (i+2<=n&&j-1>=1&&c[i+2][j-1]==0) build(x,(i+1)*n+j-1+sum);
                if (i+2<=n&&j+1<=n&&c[i+2][j+1]==0) build(x,(i+1)*n+j+1+sum);

            }
        }
    }
    t=0;
    int ans=0;
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
        {
            if (c[i][j]==0)
            {
                t++;
                if (find((i-1)*n+j)) ans++;
            }
        }
    }
    printf("%d\n",sum-ans/2);
    return 0;
}