HDU1081

1.题目描述：

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12543    Accepted Submission(s): 6018

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

 

Input

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

 

Output

Output the sum of the maximal sub-rectangle.

 

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

 

Sample Output

15

 

Source

Greater New York 2001

 

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2.题意概述：

给你一矩阵，定义矩阵的和为其所有子元素的和，求包括本身的所有子矩阵的和的最大值

3.解题思路：

暴力枚举肯定会T，因为问题不具有后效性，考虑动态规划，先预处理一下mp[i][j]表示第i行前j列元素的和，把多维转化为一维的，那么每次枚举i，j，求它们的最大连续子序列和就行，需要注意的是子元素可能有负，因此ans要设为-INF（实思路挺巧的，是一道很经典的好题）

4.AC代码：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 100100#define N 155#define eps 1e-6#define pi acos(-1.0)#define e 2.718281828459#define mod (int)1e9 + 7using namespace std;typedef long long ll;typedef unsigned long long ull;int dp[N][N];int main(){#ifndef ONLINE_JUDGE  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout);  long _begin_time = clock();#endif  int n;  while (~scanf("%d", &n))  {    memset(dp, 0, sizeof(dp));    for (int i = 1; i <= n; i++)      for (int j = 1; j <= n; j++)      {        int x;        scanf("%d", &x);        dp[i][j] += dp[i][j - 1] + x;      }    int ans = -INF;    for (int i = 1; i <= n; i++)      for (int j = i; j <= n; j++)      {        int sum = 0;        for (int k = 1; k <= n; k++)        {          if (sum < 0)            sum = 0;          sum += dp[k][j] - dp[k][i - 1];          ans = max(ans, sum);        }      }    printf("%d\n", ans);  }#ifndef ONLINE_JUDGE  long _end_time = clock();  printf("time = %ld ms.", _end_time - _begin_time);#endif  return 0;}