POJ 1050 To the Max【贪心】

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 48191 Accepted: 25509

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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

相当于二维的最大子序列，先把他降到一维再贪心或者dp

#include<iostream>#include<cstdio>#include<math.h>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<list>#include<vector>#include<sstream>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define ms(x,y)memset(x,y,sizeof(x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longconst int INF=0x7ffffff;const int M=1e4+10;int i,j,k,n,m;int q,t;int f[M];int num[M];int mp[M][M];int d[M];int main(){    while(~scanf("%d",&n))    {        for(i=0;i<n;i++)            for(j=0;j<n;j++)            scanf("%d",&mp[i][j]);        int ans=0;       for(i=0;i<n;i++){        ms(d,0);        for(j=i;j<n;j++){            for(k=0;k<n;k++){                d[k]+=mp[j][k];            }            int sum=0;            for(k=0;k<n;k++){                sum+=d[k];                ans=max(ans,sum);                if(sum<0)sum=0;            }          }       }       printf("%d\n",ans);    }    return  0;}