293 字
1 分钟
P1169 [ZJOI2007] 棋盘制作
题目
题解
悬线法
正方形最大面积可以统计最大变成的平方
最大边长等于min(r[i][j] - l[i][j] + 1, h[i][j])
锅
当i==1
时候l与r数组不要更新。。。。。
代码
#include<bits/stdc++.h>
using namespace std;const int N = 2100;int M[N][N];int zhuan[N][N];int n, m;int h[N][N], l[N][N], r[N][N];int main(){// freopen("in.in", "r", stdin);// freopen("out.out", "w", stdout); ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); cin >> n >> m; for(int i = 1;i <= n;i++) { for(int j = 1;j <= m;j++) { cin >> M[i][j]; h[i][j] = 1; l[i][j] = r[i][j] = j; } }
for(int i = 2;i <= n;i++) { for(int j = 1;j <= m;j++) { if(M[i - 1][j] != M[i][j]) { h[i][j] = h[i - 1][j] + 1; } } }
for(int i = 1;i <= n;i++) { for(int j = 2;j <= m;j++) { if(M[i][j - 1] != M[i][j]) { l[i][j] = l[i][j - 1]; } } }
for(int i = 1;i <= n;i++) { for(int j = m - 1;j >= 1;j--) { if(M[i][j + 1] != M[i][j]) { r[i][j] = r[i][j + 1]; } } } int maxsquare = 0; int maxlen = 0; for(int i = 1;i <= n;i++) { for(int j = 1;j <= m;j++) { if(M[i - 1][j] != M[i][j]) { if(i == 1) continue; l[i][j] = max(l[i][j], l[i - 1][j]); r[i][j] = min(r[i][j], r[i - 1][j]); }
int b = min(r[i][j] - l[i][j] + 1, h[i][j]); maxlen = max(maxlen, b * b); maxsquare = max(maxsquare, (r[i][j] - l[i][j] + 1) * h[i][j]); } } cout << maxlen << endl; cout << maxsquare << endl;
return 0;}
P1169 [ZJOI2007] 棋盘制作
https://blog.histcat.top/posts/p1169/