PAT T1012 Greedy Snake

直接暴力枚举，注意每次深搜完状态的还原~

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
int visit[maxn][maxn];
int N,M,x,y;
int cnt;
int maxcnt;
int X[]={,,-,};
int Y[]={,,,-};
int judge (int x,int y) {
    if (x<||x>N||y<||y>N) return ;
    if (visit[x][y]) return ;
    return ;
}
void dfs (int x,int y,int pre) {
    visit[x][y]=;
    cnt++;
    int tx=x+X[pre];
    int ty=y+Y[pre];
    if (judge(tx,ty)) dfs (tx,ty,pre);
    else {
        int flag=;
        for (int i=;i<;i++) {
            tx=x+X[i];
            ty=y+Y[i];
            if (judge(tx,ty)) flag++,dfs(tx,ty,i);
        }
        if (flag==) maxcnt=max(maxcnt,cnt);
    }
    visit[x][y]=;
    cnt--;
}
int main () {
    scanf ("%d %d",&N,&M);
    N-=;
    for (int i=;i<M;i++) {
        scanf ("%d %d",&x,&y);
        x--;
        y--;
        visit[x][y]=;
    }
    int ans=;
    int num=;
    for (int i=;i<=N;i++) {
        for (int j=;j<=N;j++) {
            if (judge(i,j)) {
                maxcnt=;
                dfs (i,j,);
                dfs (i,j,);
                dfs (i,j,);
                dfs (i,j,);
                if (maxcnt>ans) {
                    ans=maxcnt;
                    num=;
                }
                else if (maxcnt==ans) {
                    num++;
                }
            }
        }
    }
    printf ("%d %d\n",N*N-ans-M,num);
    num=;
    int ans1=;
    for (int i=;i<=N;i++) {
        for (int j=;j<=N;j++) {
            if (visit[i][j]) continue;
            visit[i][j]=;
            maxcnt=;
            for (int k=;k<=N;k++) {
                for (int w=;w<=N;w++) {
                    if (judge(k,w)) {
                        dfs (k,w,);
                        dfs (k,w,);
                        dfs (k,w,);
                        dfs (k,w,);
                    }
                }
            }
            if (maxcnt>ans1) {
                ans1=maxcnt;
                num=;
            }
            else if (maxcnt==ans1) {
                num++;
            }
            visit[i][j]=;
        }
    }
    if (ans1<ans) printf ("-1");
    else printf ("%d %d\n",N*N-M-ans1-,num);
    return ;
}