UVALive 4950 Selling Land

在维护高度递增的单调栈中，同时维护周长递增（这一点是很容易证明的），就可以用O(n^2)的复杂度解决了。

代码：

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define lson o<<1
#define rson o<<1|1
#define CLR(A, X) memset(A, X, sizeof(A))
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
const double eps = 1e-10;
int dcmp(int x){if(fabs(x)<eps) return 0; return x<0?-1:1;}
const int N = 1e3+5;
const LL MOD = 1e9+7;

char mp[N][N];
int h[N], num[2*N];
PII s[N];

int main() {
    int T;
    cin >> T;
    while(T--) {
        CLR(num, 0); CLR(h, 0);
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%s", mp[i]+1);
        for(int i = 1; i <= n; i++) {
            int top = -1;
            for(int j = 1; j <= m; j++) {
                if(mp[i][j] == '#') {
                    h[j] = 0;
                    top = -1;
                    continue;
                }
                h[j]++;
                PII x = {j, h[j]};
                while(top>=0 && s[top].se>=x.se) x.fi = s[top--].fi;
                if(top<0 || s[top].se-s[top].fi<x.se-x.fi) s[++top] = x;
                x = s[top];
                num[x.se-x.fi+j+1]++;
            }
        }
        for(int i = 1; i <= n+m; i++) if(num[i]) {
            printf("%d x %d\n", num[i], i<<1);
        }
    }
    return 0;
}