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Timus 1146. Maximum Sum

2024-08-15 21:29:55   218人阅读

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:
0−2−70
92−62
−41−41
−180−2
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

inputoutput
40 -2 -7 09 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        :guanjunCreated Time  :2016/10/7 13:50:13File 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;}

 

Timus 1146. Maximum Sum


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