HDU 4185 二分匹配 1*2方格个数

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题意：

问'##' 这样最多有几个相邻

与之前的用1*2铺矩阵题目一样，把所有非#的点改成坏点，水过

#include <iostream>#include <stdio.h>#include <string.h>#include <set>#include <queue>#include <algorithm>#include <cstdlib>#include <vector>#include <math.h>#define N 202#define M 201using namespace std;int n;int map[N][N];int lef[N*N];//lef[v]表示Y集的点v 当前连接的点  bool T[N*N];     //T[u] 表示Y集 u 是否已连接x集vector<int>G[N*N]; //匹配边  G[X集].push_back(Y集)  注意G 初始化bool match(int x){ // x和Y集 匹配 返回x点是否匹配成功for(int i=0; i<G[x].size(); i++){int v = G[x][i];if(!T[v]){T[v] = true;if(lef[v] == -1 || match( lef[v] )){lef[v] = x;return true;}}}return false;}int solve(){memset(lef, -1, sizeof(lef));int ans = 0;for(int i = 1; i<= n; i++)//X集匹配，X集点标号从 1-pn 匹配边是G[左点].size()   for(int j = 1; j<=n; j++){memset(T, 0, sizeof(T));if( match( i*M +j ) ) ans++;}return ans;}void pipei(int x1,int y1,int x2,int y2){if(map[x2][y2] || map[x1][y1])  return ;if(x1>n || x2>n || y1>n || y2>n)return ;G[ x1*M + y1 ] .push_back( x2*M + y2);}void Have_map(){int i,j;for(i=1;i<=n;i++)for(j=1;j<=n;j++)G[i*M+j].clear();for(i=1;i<=n;i++)for(j=1;j<=n;j++){pipei(i,j, i,  j+1);pipei(i,j+1, i,j);pipei(i,j, i+1,  j);pipei(i+1, j, i, j);}}char s[N];int main(){int i, j, t, Cas = 1;scanf("%d",&t);while(t--){memset(map,0,sizeof(map));scanf("%d",&n);for(i = 1; i <= n; i++){scanf("%s",s);for(j = 1; j <= n; j++)if(s[j-1] == '.')map[i][j] = 1;}Have_map();int ans = solve();printf("Case %d: %d\n", Cas++, ans>>1);}return 0;}/*16.......##....##.......#.....##......*/