HDU5556 Land of Farms(二分图 2015 合肥区域赛)

Posted 2020-08-09

容易想到将问题转化为求图的独立数问题 ，但求一般图的独立集是一个NPC问题，需要一些转化。

状态压缩，枚举每个上古农场是否选择，然后将剩下的新农场根据i + j奇偶性分为x , y集。

 

结果为 max(tot  + nx + ny - 二分图匹配数)

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 150, INF = 0x3F3F3F3F;

int mp[15][15];

int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

int cx[N], cy[N];//匹配结果
int nx, ny;//x,y集合元素个数
bool inx[N], iny[N];

bool used[N];
bool nop[15][15];

//存x到y集合的有向图
struct Node{
    int to,next;
}edge[1000];
int head[N],tot;

bool dfs(int u){
    for(int i = head[u]; ~i; i = edge[i].next){
        int v = edge[i].to;
        if(!used[v]){
            used[v] = true;
            if(cy[v] == -1 || dfs(cy[v])){
                cx[u] = v;
                cy[v] = u;
                return true;
            }
        }
    }
    return false;
}

int Hungary(){
    int ans = 0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));
    for(int i = 0; i < nx; i++){
        if(cx[i] == -1){
            memset(used, 0, sizeof(used));
            if(dfs(i)){
                ans++;
            }
        }
    }
    return ans;
}

void init(){
    memset(head, -1, sizeof(head));
    tot = 0;
}
void add(int u, int to){
    edge[tot].to=to;
    edge[tot].next=head[u];
    head[u]=tot++;
}

bool check(int n, int m, int st){
    memset(nop, 0, sizeof(nop));
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(st & (1 << mp[i][j])){
                for(int k = 0; k < 4; k++){
                    int u = i + dir[k][0];
                    int v = j + dir[k][1];
                    if((mp[u][v] != -1) && (mp[i][j] != mp[u][v]) && (st & (1 << mp[u][v]))){
                        return false;
                    }
                    nop[u][v] = 1;
                }
            }
        }
    }
    return true;

}
char str[N];
int main(){
    int t, n, m;
    cin>>t;
    int vis[N];
    for(int cas = 1; cas <= t; cas++){
        scanf("%d %d", &n, &m);
        int ans = 0;
        memset(mp, -1, sizeof(mp));
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++){
            scanf("%s", str + 1);
            for(int j = 1; j <= m; j++){
                if(str[j] == ‘.‘){
                    mp[i][j] = 10;
                }else{
                    mp[i][j] = str[j] - ‘0‘;
                    vis[str[j] - ‘0‘] = 1;
                }
            }
        }
        int cnt = 0;
        for(int i = 0; i < 10; i++){
            if(vis[i]){
                vis[i] = cnt++;
            }
        }


        for(int i = 1; i <= n; i++){
                for(int j = 1; j <= m; j++){
                    if(mp[i][j] < 10){
                        mp[i][j] = vis[mp[i][j]];
                }
            }
        }

        for(int st = 0; st < (1 << cnt); st++){
            int tot = 0;
            int tp = st;
            while(tp){
                tot += tp & 1;
                tp >>= 1;
            }

            if(!check(n, m, st)){
                continue;
            }
            init();
            nx = 0;
            ny = 0;
            memset(inx, 0, sizeof(inx));
            memset(iny, 0, sizeof(iny));
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= m; j++){
                    if(mp[i][j] < 10){
                        nop[i][j] = 1;
                    }

                    if((i + j) % 2){
                        if(mp[i][j] == 10 && !nop[i][j]){
                            iny[(i - 1) *m + j - 1] = 1;
                        }
                    }

                    if((i + j) % 2 == 0){
                        if(mp[i][j] == 10 && !nop[i][j]){
                            inx[(i - 1) *m + j - 1] = 1;
                            for(int k = 0; k < 4; k++){
                                int u = i + dir[k][0];
                                int v = j + dir[k][1];
                                if(mp[u][v] == 10 && !nop[u][v]){
                                    add((i - 1) *m + j - 1, (u - 1) *m + v - 1);
                                }


                            }

                        }

                    }

                }
            }
            for(int i = 0; i < n * m; i++){
                if(inx[i]){
                    tot++;
                    nx = i + 1;
                }
                if(iny[i]){
                    tot++;
                    ny = i + 1;
                }
            }
            ans = max(ans, tot - Hungary());
        }

        printf("Case #%d: %d\n", cas, ans);
    }

    return 0;
}