pku3023 Antenna Placement

http://acm.pku.edu.cn/JudgeOnline/problem?id=3020

题意:给出N*M大小的棋盘,棋盘以'*'与'o'组成,现在要求用1*2大小的矩形覆盖这些'*'的点,问最少要用多少个.

思想:此题与Chessboard很相近,就是把棋盘进行编号,如果某点是'*',则与其上下左右为'*'的点建立边,不过此题

建立的是双向边,找到的最大匹配数要div2,利用个数=顶点数-最大匹配div2,就可以啦!

源码:

#include<iostream>
using namespace std;
#define MaxInt 402
int xM[MaxInt],yM[MaxInt];
bool chk[MaxInt];
bool Map[MaxInt][MaxInt];
char base[42][12];
int x,y;
int n,m,k;
bool dfs(int u)
{
    int v;
    for(v=1;v<=y;v++)
    {
        if(Map[u][v]&&!chk[v])
        {
            chk[v]=true;
            if(yM[v]==-1||dfs(yM[v]))
            {
                yM[v]=u;
                xM[u]=v;
                return true;
            }
        }
    }
    return false ;
}
int main()
{
    int i,j;
    int num;
    scanf("%d",&k);
    while(k--)
    {
        scanf("%d%d",&n,&m);
            num=0;
            for(i=1;i<=n;i++)
     scanf("%s",base[i]+1);
            memset(Map,false,sizeof(Map));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    if(base[i][j]=='*')
                    {
                        ++num;
      x=(i-1)*m+j;
                        if(i>1&&base[i-1][j]=='*')
      {
                            Map[x][x-m]=true;
       Map[x-m][x]=true;
      }
                        if(i<n&&base[i+1][j]=='*')
      {
                            Map[x][x+m]=true;
       Map[x+m][x]=true;
      }
                        if(j>1&&base[i][j-1]=='*')
      {
       Map[x][x-1]=true;
       Map[x-1][x]=true;
      }
                        if(j<m&&base[i][j+1]=='*')
      {
                            Map[x][x+1]=true;
       Map[x+1][x]=true;
      }
                    }
                }
            }
            memset(xM,0xff,sizeof(xM));
            memset(yM,0xff,sizeof(yM));
            j=0;
   x=y=n*m;
            for(i=1;i<=x;i++)
            {
                if(xM[i]==-1)
                {
                    memset(chk,false,sizeof(chk));
                    if(dfs(i))
                        j++;
                }
            }
            printf("%d/n",num-j/2);
    }
    return 0;
}