Timus 1146. Maximum Sum

1146. Maximum Sum

Time limit: 0.5 second
Memory limit: 64 MB

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the sum of 15.

Input

The input consists of an N × N array of integers. The input begins with a single positive integerN on a line by itself indicating the size of the square two dimensional array. This is followed byN 2 integers separated by white-space (newlines and spaces). These N 2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

Sample

input	output
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2	15

最大子矩阵。很经典的问题哈哈

压缩 然后最大连续子序列  dp[i]=dp[i-1]<0?a[i]:dp[i-1]+a[i]

一开始压缩的时候没用前缀和，n^4 貌似过不了，后来用前缀和优化到n^3  

下面代码中dp 的空间也可以优化，这里没有优化.

/* ***********************************************
Author        :guanjun
Created Time  :2016/10/7 13:50:13
File Name     :timus1146.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int a[110][110],n;
int sum[110][110];
int dp[110];
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    while(scanf("%d",&n)!=EOF){
        cle(sum);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&a[i][j]);
                sum[i][j]=sum[i][j-1]+a[i][j];
            }
        }
        int Max=-INF;
        //dp  求最大连续子序列 dp[i]代表以i为结尾的最大连续子序列的长度
        for(int i=1;i<=n;i++){
            for(int j=1;j<=i;j++){
                cle(dp);
                for(int k=1;k<=n;k++){
                    int tmp=sum[k][i]-sum[k][j-1];
                    if(dp[k-1]<0){
                        dp[k]=tmp;
                    }
                    else dp[k]=tmp+dp[k-1];
                    Max=max(dp[k],Max);
                }
            }
        }
        cout<<Max<<endl;
    }
    return 0;
}