poj3020 Antenna Placement

只能说构图是痛苦的。。。。。完全没思路，思路在下面。。。。

无向二分图的最小路径覆盖 = 顶点数 – 最大二分匹配数/2

http://user.qzone.qq.com/289065406/blog/1299322779

code

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 502;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
const double eps= 1e-7;
using namespace std;
int N,M,V1,V2,cnt,ans;//V1,V2是二分图的两个顶点集合
int city[maxn][maxn];
bool vis[maxn];//记录一边的顶点是否搜索过
bool Graph[maxn][maxn];
int link[maxn];//记录连接这个y顶点的x点
bool dfs(int x){
    FOR(i,1,V2){
        if(!vis[i] && Graph[x][i]){
            vis[i]=true;
            if(link[i]==0 || dfs(link[i])){
                link[i]=x;
                return true;
            }
        }
    }
    return false;
}
void getGraph(){
    FOR(i,1,N) FOR(j,1,M){
        if(city[i][j]){
            FF(k,4){
                int nx=i+dx[k];
                int ny=j+dy[k];
                if(city[nx][ny]){
                    Graph[city[i][j]][city[nx][ny]]=true;
                }
            }
        }
    }
}
int main()
{
    int cas;
    SD(cas);
    char ch;
    while(cas--){
        cnt=ans=0;
        SET(city,0);SET(Graph,0);SET(link,0);
        SD(N);SD(M);
        FOR(i,1,N) FOR(j,1,M){
            scanf(" %c",&ch);
            if(ch=='*'){
                city[i][j]=++cnt;//代表这个点有城市
            }
        }
        getGraph();
        V1=V2=cnt;
        FOR(i,1,V1){
            SET(vis,false);
            if(dfs(i)) ans++;
        }
        PD(cnt-ans/2);LN;
    }
    return 0;
}