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  • Luogu P1169 [ZJOI2007]棋盘制作

    gate

    悬线法

    用于求极大子矩阵

    (l[i][j]:(i,j))能扩展到的最左边的点
    (r[i][j]:(i,j))能扩展到的最右边的点
    (up[i][j]:(i,j))能扩展到的最上边的长度

    预处理:(如果合法)
    (l[i][j] = l[i][j-1])
    (r[i][j] = r[i][j+1])

    状态转移方程:(如果合法)
    (l[i][j] = max(l[i][j],l[i-1][j]))
    (r[i][j] = min(r[i][j],r[i-1][j]))
    (up[i][j] = up[i-1][j]+1)

    也就是,如果这一点能从上面的点转移过来,则左右端点取较小的。
    像这样:

    代码如下

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<cstring>
    #define MogeKo qwq
    using namespace std;
    
    const int maxn = 2005;
    int n,m,wid,ans1,ans2;
    int a[maxn][maxn],l[maxn][maxn],r[maxn][maxn],up[maxn][maxn];
    
    int main() {
    	scanf("%d%d",&n,&m);
    	for(int i = 1; i <= n; i++)
    		for(int j = 1; j <= m; j++) {
    			scanf("%d",&a[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(a[i][j] != a[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(a[i][j] != a[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 && a[i][j] != a[i-1][j]) {
    				l[i][j] = max(l[i][j],l[i-1][j]);
    				r[i][j] = min(r[i][j],r[i-1][j]);
    				up[i][j] = up[i-1][j]+1;
    			}
    			wid = min(up[i][j],r[i][j]-l[i][j]+1);
    			ans1 = max(ans1,wid*wid);
    			ans2 = max(ans2,up[i][j]*(r[i][j]-l[i][j]+1));
    		}
    	printf("%d
    %d",ans1,ans2);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mogeko/p/12969541.html
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