hdu5556 Land of Farms

我对于题目的一种理解
改造农场
1.建新农场 在空的点选
2.重建旧农场 选一个点属于这个农场的地方都要选

最后的农场都不能相连
所以枚举旧农场的个数并进行二分图匹配

#include<bits/stdc++.h>
using namespace std;
int N,M;

char mp[12][12];
vector<pair<int,int> > farm[12];
int has[12];
int dir[5][5] = { {1,0}, {0,1}, {-1,0}, {0,-1} };
int vis[12][12];
int ok(int x,int y) {
    if(x >= 1 && x <= N && y >= 1 && y <= M) return 1;
    else return 0;
}
int judge(int x, int y, int ty) {
    for(int i = 0; i < 4; ++i) {
        int t1 = x + dir[i][0]; int t2 = y + dir[i][1];
        if( ok(t1,t2) && mp[t1][t2] != '.' && mp[t1][t2] != ty+'0' && has[mp[t1][t2] - '0'] ) return 0;
        else if( ok(t1,t2) && mp[t1][t2] == '.' ) vis[t1][t2] = 1;
    }   
    return 1;
}

int linker[12][12];
int used[12][12];
int dfs(int x, int y) {
    if( (x+y)%2 == 1 ) return 0;
//  printf("%d %d
",x,y);
    for(int i = 0; i < 4; ++i) {
        int t1 = x + dir[i][0]; int t2 = y + dir[i][1];
        if( ok(t1,t2) && !used[t1][t2] && !vis[t1][t2]) {
            used[t1][t2] = 1;
            if( linker[t1][t2] == -1 || dfs(linker[t1][t2]/11, linker[t1][t2]%11) ) {
                linker[t1][t2] = x*11+y; 
                return 1;
            }
        }
    }
    return 0;
}
int ccc = 0;
int main(){
    int T; scanf("%d",&T); int ca = 0;
    while(T--) {
        ccc = 0;
        scanf("%d %d",&N,&M);
        int ans = -1;
        for(int i = 0; i <= 10; ++i) farm[i].clear();
        for(int i = 1; i <= N; ++i) scanf("%s",mp[i]+1);

        for(int i = 1; i <= N; ++i) {
            for(int j = 1; j <= M; ++j) {
                if(mp[i][j] != '.') {
                    farm[mp[i][j] - '0'].push_back({i,j});
                } 
            }
        }

        for(int i = 0; i < (1<<10); ++i) {
            int cc = 0;
            memset(has,0,sizeof(has));
            for(int j = 0; j < 10; ++j) {
                if( (i& (1<<j)) && (int)farm[j].size() ) {
                    has[j] ++; cc ++;
                }
            }
            if(!cc && i!=0) continue;
            memset(vis,0,sizeof(vis));
            int fl = 1;
            for(int j = 0; j < 10 && fl; ++j) {
                if(has[j]) {
                    for(int k = 0; k < (int)farm[j].size() && fl; ++k) {
                        fl = judge(farm[j][k].first, farm[j][k].second, j);
                    }
                }
            }
            if(!fl) continue;

            int tot = cc;
            for(int j = 1; j <= N; ++j)
                for(int k = 1; k <= M; ++k) {
                    if(mp[j][k] != '.') {
                        vis[j][k] = 1;
                    }else if(!vis[j][k]) tot ++;
                }
        /*  
            if(ccc < 10) {
                for(int j = 0; j < 10; ++j) if(has[j]) printf("%d ",j); printf("
");
                for(int j = 1; j <= N; ++j) {
                    for(int k = 1; k <= M; ++k) {
                        printf("%d ",vis[j][k]);
                    }
                    printf("
");
                }
                ccc ++;

            }
            */


            int res = 0;
            memset(linker, -1, sizeof(linker));
            for(int j = 1; j <= N; ++j) {
                for(int k = 1; k <= M; ++k) {
                    if(vis[j][k]) continue;
                    memset(used,0,sizeof(used));
                    if(dfs(j,k)) res ++;
                }
            }

            ans = max(tot - res, ans);
        }

        printf("Case #%d: %d
", ++ca,ans);
    }
    return 0;
}