zoj1074 To the Max

Problem

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.

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.

Example

Input

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

Output

15 

这题是序列最大和的升级版，我们可以枚举两行p，q，然后把p，q行每一列的元素的和都加起来，放到sum[]这个数组中，然后求出sum[]数组的序列最大和就行了。复杂度是O(n^3).

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 106
ll a[maxn][maxn],sum[maxn];
ll f(ll c[],ll n)
{
    ll i,j;
    ll b[maxn];
    b[0]=-inf;
    ll maxx=-inf;
    for(i=1;i<=n;i++){
        b[i]=max(c[i],b[i-1]+c[i]);
        maxx=max(maxx,b[i]);
    }
    return maxx;
}




int main()
{
    ll n,m,i,j,maxx,k;
    while(scanf("%lld",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%lld",&a[i][j]);
            }
        }
        maxx=-inf;
        for(i=1;i<=n;i++){
            memset(sum,0,sizeof(sum));
            for(j=i;j<=n;j++){
                for(k=1;k<=n;k++){
                    sum[k]+=a[j][k];
                }
                maxx=max(maxx,f(sum,n));
            }

        }
        printf("%lld
",maxx);
    }
    return 0;
}