Antenna Placement

Antenna Placement

题目描述：

N * M的棋盘中，'o'代表障碍物，'*'代表空位，你可以使用1 * 2的多米诺骨牌放进空格，米牌可以重叠，也可以放在障碍物上，问最少使用多少牌使得放满所有空位

思路：

和上次那个棋盘问题很相似了，不过这次可以重叠了，那我们只需要计算出最大匹配值，然后就可以得出单放的空位是多少，最后需要放的牌的数量为：，num是空位的多少，pi是最大匹配数

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<sstream>
#include<cstring>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define MAX  100 + 50
#define mod 1000000007
#define lowbit(x) (x & (-x))
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n",n, m)
#define sddd(n,m,z) scanf("%d %d %d",&n,&m,&z)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define mem(a,b) memset((a),(b),sizeof(a))
typedef  long long ll ;
typedef unsigned long long ull;
//不开longlong见祖宗!提交不看数据范围见祖宗!
inline int IntRead(){char ch = getchar();int s = 0, w = 1;while(ch < '0' || ch > '9'){if(ch == '-') w = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){s = s * 10 + ch - '0';ch = getchar();}return s * w;}

int t;
int n, m;
int white, black;
char mp[MAX][MAX];
int tr[MAX][MAX];
bool p[MAX][MAX];
vector<int>v[10005];
bool vis[10005];
int link[10005];

bool dfs(int x){
    for(int i = 0; i < v[x].size(); ++i){
        int k = v[x][i];
        if(!vis[k]){
            vis[k] = 1;
            if(!link[k] || dfs(link[k])){
                link[k] = x;
                return true;
            }
        }
    }
    return false;
}

void work(){
    int ans = 0;
    for(int i = 1; i <= white; ++i){
        mem(vis, 0);
        if(dfs(i))++ans;
    }
    cout<<white + black - ans<<endl;

}

void init(){
    white = black = 0;
    mem(p, 0);
    mem(link, 0);
    mem(tr, 0);
    for(int i = 0; i <= n * m; ++i)v[i].clear();
}

int main(){
    sd(t);
    while (t--) {
        sdd(n, m);
        init();
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                cin>>mp[i][j];
                if(mp[i][j] == '*')p[i][j] = 1;
            }
        }
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                if(p[i][j]){
                    if((i + j) & 1)tr[i][j] = ++white;
                    else tr[i][j] = ++black;
                }
            }
        }
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                if(p[i][j] && (i + j) & 1){
                    if(tr[i][j - 1])v[tr[i][j]].push_back(tr[i][j - 1]);
                    if(tr[i][j + 1])v[tr[i][j]].push_back(tr[i][j + 1]);
                    if(tr[i + 1][j])v[tr[i][j]].push_back(tr[i + 1][j]);
                    if(tr[i - 1][j])v[tr[i][j]].push_back(tr[i - 1][j]);
                }
            }
        }
        work();

    }
    return 0;
}