ZOJ - 1654Place the Robots 建图+二分图最大匹配

ZOJ - 1654Place the Robots 建图+二分图最大匹配

https://vjudge.net/problem/ZOJ-1654

分析： 每一个空地使得这个位置相邻的空地不能被选择，将空地分为一个一个块，即求最大点独立集，即最大匹配

想了好久没想到，一开始直接对点建图，死的不知道有多惨，愣是想了一个小时不知道怎么建图

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2500+7;
const int INF=1e9;
const ll INFF=1e18;
int mapp[maxn][maxn],hhh[51][51],ggg[51][51],n,m,xx,yy,t;
char s[51][51];
int linker[maxn];
bool vis[maxn];
bool dfs(int x)
{
    for (int i=1;i<=yy;i++)
    {
        if (mapp[x][i]&&!vis[i])
        {
            vis[i]=true;
            if (linker[i]==-1||dfs(linker[i]))
            {
                linker[i]=x;
                return true;
            }
        }
    }
    return false;
}
int count_()
{
    int res=0;
    for (int i=1;i<=xx;i++)
    {
        mem(vis,false);
        if (dfs(i))res++;
    }
    return res;
}
int main()
{
    scanf("%d",&t);
    rep(q,1,t)
    {
        mem(linker,-1);mem(mapp,0);mem(hhh,0);mem(ggg,0);
        scanf("%d%d",&m,&n);
        repp(i,0,m)scanf("%s",s[i]);
        xx=0;yy=0;
        for (int i=0;i<m;i++)
        {
            int k=0;
            while(k<n)
            {
                if (s[i][k]=='o')
                {
                    xx++;
                    while(k<n&&s[i][k]!='#')
                    {
                        hhh[i][k]=xx;
                        k++;
                    }
                }
                else k++;
            }
        }
        for (int i=0;i<n;i++)
        {
            int k=0;
            while(k<m)
            {
                if (s[k][i]=='o')
                {
                    yy++;
                    while(k<m&&s[k][i]!='#')
                    {
                        ggg[k][i]=yy;
                        k++;
                    }
                }
                else k++;
            }
        }
        repp(i,0,m)
            repp(j,0,n)
                if (s[i][j]=='o')mapp[hhh[i][j]][ggg[i][j]]=1;
        printf("Case :%d\n",q);
        W(count_());
    }
    return 0;
}