codeforces 15d

题意：定一个n*m的区域，并给出该区域各个位置的地基高度，要在该区域上建a*b的房子若干栋，而建造每栋房子的地基高度需要相同。因此在一块区域内建房子时若地基不同，需要将高的地方铲成低的地方的高度，然后需要两者高度差的花费。要求每栋房子花费尽量小（单个花费最小，不是整体花费最小），花费相同时左上角的优先输出
 

	#include <bits/stdc++.h>
	using namespace std;
	
	typedef long long LL;
	const int N = 1005;
	
	LL high[1007][1007];
	bool vis[1007][1007];
	LL dp[1005][20];
	LL dp2[N][20];
	LL f[N][N];
	LL sum[N][N];
	
	struct Node{
	    int x, y;
	    LL cost;
	    Node() {}
	    Node(int xx, int yy, LL c) : x(xx), y(yy), cost(c) {}
	    friend bool operator < (Node aa, Node bb){
	        if(aa. cost == bb. cost){
	            if(aa. x == bb. x)
	                return aa. y < bb. y;
	            return aa. x < bb. x;
	        }
	        return aa. cost < bb. cost;
	    }
	}d[1000005];
	int n, m, a, b;
	vector <Node> ans;
	
	int main(){
	    while(scanf("%d %d %d %d", &n, &m, &a, &b) != EOF){
	        memset(sum, 0, sizeof sum);
	        memset(f, 0, sizeof f);
	        for(int i = 1; i <= n; i ++)
	            for(int j = 1; j <= m; j ++){
	                scanf("%I64d", &high[i][j]);
	                sum[i][j] = (sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1]) + high[i][j];
	            }
	        for(int i = 1; i <= n; i ++){
	            for(int j = 1; j <= m; j ++)
	                dp[j][0] = high[i][j];
	            for(int k = 1; (1 << k) <= b; k ++)
	                for(int j = 1; j + (1 << k) - 1 <= m; j ++)
	                    dp[j][k] = min(dp[j][k - 1], dp[j + (1 << (k - 1))][k - 1]);
	            int k = log(b * 1.0) / log(2.0);
	            for(int j = 1; j <= m - b + 1; j ++){
	                f[i][j] = min(dp[j][k], dp[j + b - (1 << k)][k]);
	            }
	        }
	        for(int i = 1; i <= m - b + 1; i ++){
	            for(int j = 1; j <= n; j ++){
	                dp[j][0] = f[j][i];
	            }
	
	            for(int k = 1; (1 << k) <= a; k ++)
	                for(int j = 1; j + (1 << k) - 1 <= n; j ++){
	                    dp[j][k] = min(dp[j][k - 1], dp[j + (1 << (k - 1))][k - 1]);
	                }
	            int k = log(a * 1.0) / log(2.0);
	            for(int j = 1; j <= (n - a + 1); j ++){
	                f[j][i] = min(dp[j][k], dp[j + a - (1 << k)][k]);
	            }
	        }
	        int cnt = 0;
	        for(int i = 1; i <= n - a + 1; i ++)
	            for(int j = 1; j <= m - b + 1; j ++){
	            LL tmp = sum[i + a - 1][j + b - 1] - sum[i + a - 1][j - 1] - sum[i - 1][j + b - 1] + sum[i - 1][j - 1] - f[i][j] * a * b;
	            d[cnt ++] = Node(i, j, tmp);
	        }
	        sort(d, d + cnt);
	        ans. clear();
	        memset(vis, false, sizeof vis);
	        for(int i = 0; i < cnt; i ++){
	            if(!vis[d[i]. x][d[i]. y] && !vis[d[i]. x + a - 1][d[i]. y] && !vis[d[i]. x][d[i]. y + b - 1] && !vis[d[i]. x + a - 1][d[i]. y + b - 1]){
	                for(int j = d[i]. x; j < d[i]. x + a; j ++)
	                    for(int k = d[i]. y; k < d[i]. y + b; k ++)
	                    vis[j][k] = true;
	                ans. push_back(d[i]);
	            }
	        }
	        printf("%d\n", ans. size());
	        for(int i = 0; i < ans. size(); i ++){
	            printf("%d %d %I64d\n", ans[i]. x, ans[i]. y, ans[i]. cost);
	        }
	    }
	    return 0;
	}