ural 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

 

Tags: dynamic programming  (hide tags for unsolved problems)

Difficulty: 97

 

题意：求最大的子矩阵

分析：很经典的题。

知道最大子段和的做法。

然后枚举矩阵的上下界，按照最大子段和的做法做。

 

 

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 int n, arr[N][N];
 int sum[N][N], p[N];

 inline void Input()
 {
     scanf("%d", &n);
     for(int i = ; i <= n; i++)
         for(int j = ; j <= n; j++) scanf("%d", &arr[i][j]);
 }

 inline int Work(int *arr)
 {
     int ret = arr[], cnt = arr[];
     for(int i = ; i <= n; i++)
     {
         if(cnt < ) cnt = arr[i];
         else cnt += arr[i];
         ret = max(ret, cnt);
     }
     return ret;
 }

 inline void Solve()
 {
     int ans = -INF;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= n; j++)
             sum[i][j] = sum[i - ][j] + arr[i][j];
     for(int i = ; i <= n; i++)
         for(int j = i; j <= n; j++)
         {
             for(int k = ; k <= n; k++)
                 p[k] = sum[j][k] - sum[i - ][k];
             int cnt = Work(p);
             /*if(ans < cnt)
             {
                 ans = cnt;
                 printf("%d %d %d\n", ans, i, j);
             }*/
             ans = max(ans, cnt);
         }

     cout << ans << endl;
 }

 int main()
 {
     freopen("a.in", "r", stdin);
     Input();
     Solve();
     return ;
 }