hdu5556 Land of Farms

来源：互联网发布：安卓开发java基础编辑：程序博客网时间：2024/06/06 03:39
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\n",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("\n");                for(int j = 1; j <= N; ++j) {                    for(int k = 1; k <= M; ++k) {                        printf("%d ",vis[j][k]);                    }                    printf("\n");                }                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\n", ++ca,ans);    }    return 0;}
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