HDU4185Oil Skimming（行列匹配||棋盘匹配||黑白染色||1X2矩形覆盖）

本文主要是介绍HDU4185Oil Skimming（行列匹配||棋盘匹配||黑白染色||1X2矩形覆盖），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题意：找出最多的形如“##”横着竖着都可以，明显的1X2矩形覆盖，直接按坐标和的奇偶来分为二分图。


#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<vector>
#include<cstdlib>
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define cl(a,b) memset(a,b,sizeof(a));
#define LL long long
#define P pair<int,int>
#define X first
#define Y second
#define pb push_back
#define fread(zcc)  freopen(zcc,"r",stdin)
#define fwrite(zcc) freopen(zcc,"w",stdout)
using namespace std;
const int maxn=610;
const int inf=999999;vector<int> G[maxn*maxn];
int matching[maxn*maxn];//因为对坐标是按行展开的编号，所以这里都要对应变大
bool vis[maxn*maxn];
int Nx;
bool dfs(int u){int N=G[u].size();for(int i=0;i<N;i++){int v=G[u][i];if(vis[v])continue;vis[v]=true;if(matching[v]==-1||dfs(matching[v])){matching[v]=u;return true;}}return false;
}
int hungar(){int ans=0;cl(matching,-1);for(int i=0;i<Nx;i++){cl(vis,false);if(dfs(i))ans++;}return ans;
}
char s[maxn][maxn];
int dir[][2]={1,0,0,1,-1,0,0,-1};
int main(){int T,cas=1;scanf("%d",&T);while(T--){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",s[i]);//图要全部读完，再去建立图}for(int i=0;i<n;i++){//建立二分图for(int j=0;j<n;j++)if((i+j)%2==0&&s[i][j]=='#'){for(int k=0;k<4;k++){int xx=dir[k][0]+i;int yy=dir[k][1]+j;if(xx>=0&&xx<n&&yy>=0&&yy<n&&s[xx][yy]=='#'){G[i*n+j].pb(xx*n+yy);}}}}Nx=n*n;printf("Case %d: %d\n",cas++,hungar());for(int i=0;i<=n*n;i++)G[i].clear();}return 0;
}