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poj 1081 To The Max

2024-09-13 17:54:48   218人阅读

                                                               To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12697    Accepted Submission(s): 6090


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

 

Sample Output
15
 
题意：二维的矩阵，从中找到一个子矩阵，使得子矩阵的和最大。
思路：可以先考虑一维的情况，一维时即数列，求数列中连续子列的和的最大值，做法就是在线处理，从头到尾一个一个元素考虑并累加过去，记当前累加值为sum，若累加的时候当前sum值小于0了，那么舍弃前面的累加列，sum更新为0，并且从下一个位置
的元素重新开始累加，途中不断的更新sum，找出最大的sum值即可，二维的情况可以看作一维的延伸情况，如果把列固定住(即选取矩阵连续的几列并固定，先算好每一行的这几列的和值)，此时纵向的从上到下累加就可以看成是一维情况下的累加，算法类同。
AC代码：
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<functional>
using namespace std;
const int N_MAX= 100+2;
int a[N_MAX][N_MAX];
int sum[N_MAX][N_MAX];
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        memset(sum,0,sizeof(sum));
        memset(a, 0,sizeof(a));
        for (int i = 0; i < n; i++) {
            for (int j = 1; j <= n; j++) {
                scanf("%d", &a[i][j]);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 1; j <= n; j++) {
                 sum[i][j] =sum[i][j-1]+ a[i][j];
            }
        }
        
        int res = -INT_MAX;
        for (int i = 0; i < n; i++) {//固定i,j
            for (int j = i+1; j <= n; j++) {
                int S = 0;
                for (int k = 0; k < n; k++) {
                    S += sum[k][j]-sum[k][i-1];//累加上闭区间[i,j]值的和
                    if (S > res)
                        res = S;
                    if (S < 0)S = 0;
                    
                }
            }
        }
        printf("%d\n",res);

    }
return 0;
}

 

 
思路
 

poj 1081 To The Max


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