HNU13028Attacking rooks (二分匹配，一行变多行，一列变多列)

Professor Anand presented the N-rooks problem to his students. Since rooks only attack each other when they share a row or column, they soon discovered that the problem can be easily solved by placing the rooks along a main diagonal of the board. So, the professor
decided to complicate the problem by adding some pawns to the board. In a board with pawns, two rooks attack each other if and only if they share a row or column and there is no pawn placed between them. Besides, pawns occupy some squares, which gives an additional
restriction on which squares the rooks may be placed on.

Given the size of the board and the location of the pawns, tell Professor Anand the maximum number of rooks that can be placed on empty squares such that no two of them attack each other.

Sample input 1
5
X....
X....
..X..
.X...
....X

Sample input 2
4
....
.X..
....
....

Sample input 3
1
X

Sample output 1
7

Sample output 2
5

Sample output 3
0

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
using namespace std;
vector<int>map[10005];
int match[10005],vist[10005];
int find(int x)
{
    int len=map[x].size();
    for(int i=0;i<len;i++)
    if(vist[map[x][i]]==0)
    {
        vist[map[x][i]]=1;
        if(match[map[x][i]]==0||find(match[map[x][i]]))
        {
            match[map[x][i]]=x;
            return 1;
        }
    }
    return 0;
}
int main()
{
    int n,rn,ln,t=0,mpr[105][105],mpl[105][105];
    char c;
    while(scanf("%d",&n)>0)
    {
        for(int i=1;i<=n;i++)
        {
            getchar();
            for(int j=1;j<=n;j++)
            {
                mpr[i][j]=mpl[i][j]=0;
                scanf("%c",&c);
                if(c=='X')
                mpr[i][j]=mpl[i][j]=-1;
            }
        }
        rn=0;
        for(int i=1;i<=n;i++)//一行变多行
        for(int j=1;j<=n;j++)
        {
            while(mpr[i][j]==-1&&j<=n)j++;
            rn++;
            while(mpr[i][j]!=-1&&j<=n)
            {
                mpr[i][j]=rn; j++;
            }
        }
        ln=0;
        for(int j=1;j<=n;j++)//一列变多列
        for(int i=1;i<=n;i++)
        {
            while(mpl[i][j]==-1&&i<=n)i++;
            ln++;
            while(mpl[i][j]!=-1&&i<=n)
            {
                mpl[i][j]=ln; i++;
            }
        }
        for(int i=1;i<=rn;i++)
        map[i].clear();
         for(int i=1;i<=n;i++)//行列建图
        for(int j=1;j<=n;j++)
        if(mpr[i][j]!=-1)
        map[mpr[i][j]].push_back(mpl[i][j]);

        memset(match,0,sizeof(match));
        int ans=0;
        for(int i=1;i<=rn;i++)
        {
            for(int j=0;j<=ln;j++)
            vist[j]=0;
            ans+=find(i);
        }

        printf("%d\n",ans);
    }
}