【二维树状数组】hdu 1559 最大子矩阵

最大子矩阵

Time Limit: 30000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5040    Accepted Submission(s): 2641

Problem Description

给你一个m×n的整数矩阵，在上面找一个x×y的子矩阵，使子矩阵中所有元素的和最大。

 

Input

输入数据的第一行为一个正整数T，表示有T组测试数据。每一组测试数据的第一行为四个正整数m,n,x,y（0<m,n<1000 AND 0<x<=m AND 0<y<=n），表示给定的矩形有m行n列。接下来这个矩阵，有m行，每行有n个不大于1000的正整数。

 

Output

对于每组数据，输出一个整数，表示子矩阵的最大和。

 

Sample Input

14 5 2 23 361 649 676 588992 762 156 993 169662 34 638 89 543525 165 254 809 280

 

Sample Output

2474

 
///AC代码

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int arr[1010][1010], m, n, x, y;int lowbit(int x){    return x & -x;}///二维树状数组inline void add(int x, int y, int tt)        ///单点修改{    for (int i = x; i < m; i += lowbit(i))        for (int j = y; j < n; j += lowbit(j))        {            arr[i][j] += tt;        }}inline int sum(int x, int y){    int res = 0;    for (int i = x; i; i -= lowbit(i))        for (int j = y; j; j -= lowbit(j))        {            res += arr[i][j];        }    return res;}inline int query(int L, int B, int R, int T) ///矩阵求和{    return sum(R, T) + sum(L - 1, B - 1) - sum(R, B - 1) - sum(L - 1, T);}int main(){    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d%d%d", &m, &n, &x, &y);        for (int i = 1; i <= m; i++)        {            for (int j = 1; j <= n; j++)            {                arr[i][j] = 0;            }        }        int k, ans = 0, gg;        for (int i = 1; i <= m; i++)        {            for (int j = 1; j <= n; j++)            {                scanf("%d", &k);                add(i, j, k);            }        }        for (int i = 1; i + x - 1 <= m; i++)        {            for (int j = 1; j + y - 1 <= n; j++)            {                gg = query(i, j, i + x - 1, j + y - 1);                ans = max(ans, gg);            }        }        printf("%d\n", ans);    }    return 0;}