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poj 1050(矩阵求和问题dp)

时间:2016-02-14 23:38:47      阅读:1152      评论:0      收藏:0      [点我收藏+]
To the Max
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 44765   Accepted: 23700

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

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

8  0 -2

Sample Output

15

Source

 
思路:将其化简成子段求和问题:将第i行到第j行的数累加成一维数列,在用子段求和求解。
   
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <cstdio>
 5 #include <string>
 6 #include <cstring>
 7 #include <vector>
 8 #include <queue>
 9 #include <stack>
10 #include <set>
11 #define INF 0x3f3f3f3f
12 #define INFL 0x3f3f3f3f3f3f3f3f
13 #define zero_(x,y) memset(x , y , sizeof(x))
14 #define zero(x) memset(x , 0 , sizeof(x))
15 #define MAX(x) memset(x , 0x3f ,sizeof(x))
16 using namespace std ;
17 #define N 505
18 typedef long long LL ;
19 LL dp[N],a[N][N],s[N][N];
20 int main(){
21     //freopen("in.txt","r",stdin);
22     int n;
23     scanf("%d",&n);
24     zero(dp);zero(a);zero(s);
25     for(int i=1;i<=n;i++){
26         for(int j=1;j<=n;j++){
27             scanf("%I64d",&a[i][j]);
28         }
29     }
30     for(int i=1;i<=n;i++){
31         for(int j=1;j<=n;j++){
32             s[j][i]=s[j-1][i]+a[j][i];
33            
34         }
35     }
36    
37     LL sum=0,b=0;
38     for(int i=0;i<=n;i++){
39         for(int j=i+1;j<=n;j++){
40                 b=0;
41             for(int k=1;k<=n;k++){
42                 
43                 if(b>0)
44                     b+=(s[j][k]-s[i][k]);
45                 else
46                     b=s[j][k]-s[i][k];
47                 if(b>sum)
48                     sum=b;
49                 
50             }
52         }
53     }
54     printf("%I64d\n",sum);
55     return 0;
56 }

 

poj 1050(矩阵求和问题dp)

原文:http://www.cnblogs.com/yoyo-sincerely/p/5093805.html

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