bzoj 1057 (悬线法求最大子矩阵)

bzoj 1057

悬线法求最大子矩阵

#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
int s[N][N], l[N][N], r[N][N], up[N][N];
int n, m, ans1, ans2;
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt", "r", stdin);
#endif //ONLINE_JUDGE
    while(~scanf("%d%d", &n, &m)){
        ans1 = ans2 = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                scanf("%d", &s[i][j]);
                l[i][j] = r[i][j] = j;
                up[i][j] = 1;
            }
        }
        for(int i = 1; i <= n; i++){
            for(int j = 2; j <= m; j++){
                if(s[i][j] == 1 - s[i][j - 1]){
                    l[i][j] = l[i][j - 1];
                }
            }
        }
        for(int i = 1; i <= n; i++){
            for(int j = m - 1; j >= 1; j--){
                if(s[i][j] == 1 - s[i][j + 1]){
                    r[i][j] = r[i][j + 1];
                }
            }
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(i > 1 && s[i][j] == 1 - s[i - 1][j]){
                    up[i][j] = up[i - 1][j] + 1;
                    l[i][j] = max(l[i][j], l[i - 1][j]);
                    r[i][j] = min(r[i][j], r[i - 1][j]);
                }
                int t = r[i][j] - l[i][j] + 1;
                int tt = min(t, up[i][j]);
                ans1 = max(ans1, tt*tt);
                ans2 = max(ans2, t * up[i][j]);
            }
        }
        printf("%d\n%d\n", ans1, ans2);
    }
    return 0;
}

posted on 2019-03-25 20:31  坤sir  阅读(...)  评论(... 编辑 收藏

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