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:

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

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。

  1 /**
  2 Create By yzx - stupidboy
  3 */
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cmath>
  8 #include <deque>
  9 #include <vector>
 10 #include <queue>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <map>
 14 #include <set>
 15 #include <ctime>
 16 #include <iomanip>
 17 using namespace std;
 18 typedef long long LL;
 19 typedef double DB;
 20 #define MIT (2147483647)
 21 #define INF (1000000001)
 22 #define MLL (1000000000000000001LL)
 23 #define sz(x) ((int) (x).size())
 24 #define clr(x, y) memset(x, y, sizeof(x))
 25 #define puf push_front
 26 #define pub push_back
 27 #define pof pop_front
 28 #define pob pop_back
 29 #define ft first
 30 #define sd second
 31 #define mk make_pair
 32
 33 inline int Getint()
 34 {
 35     int Ret = 0;
 36     char Ch = ‘ ‘;
 37     bool Flag = 0;
 38     while(!(Ch >= ‘0‘ && Ch <= ‘9‘))
 39     {
 40         if(Ch == ‘-‘) Flag ^= 1;
 41         Ch = getchar();
 42     }
 43     while(Ch >= ‘0‘ && Ch <= ‘9‘)
 44     {
 45         Ret = Ret * 10 + Ch - ‘0‘;
 46         Ch = getchar();
 47     }
 48     return Flag ? -Ret : Ret;
 49 }
 50
 51 const int N = 110;
 52 int n, arr[N][N];
 53 int sum[N][N], p[N];
 54
 55 inline void Input()
 56 {
 57     scanf("%d", &n);
 58     for(int i = 1; i <= n; i++)
 59         for(int j = 1; j <= n; j++) scanf("%d", &arr[i][j]);
 60 }
 61
 62 inline int Work(int *arr)
 63 {
 64     int ret = arr[1], cnt = arr[1];
 65     for(int i = 2; i <= n; i++)
 66     {
 67         if(cnt < 0) cnt = arr[i];
 68         else cnt += arr[i];
 69         ret = max(ret, cnt);
 70     }
 71     return ret;
 72 }
 73
 74 inline void Solve()
 75 {
 76     int ans = -INF;
 77     for(int i = 1; i <= n; i++)
 78         for(int j = 1; j <= n; j++)
 79             sum[i][j] = sum[i - 1][j] + arr[i][j];
 80     for(int i = 1; i <= n; i++)
 81         for(int j = i; j <= n; j++)
 82         {
 83             for(int k = 1; k <= n; k++)
 84                 p[k] = sum[j][k] - sum[i - 1][k];
 85             int cnt = Work(p);
 86             /*if(ans < cnt)
 87             {
 88                 ans = cnt;
 89                 printf("%d %d %d\n", ans, i, j);
 90             }*/
 91             ans = max(ans, cnt);
 92         }
 93
 94     cout << ans << endl;
 95 }
 96
 97 int main()
 98 {
 99     freopen("a.in", "r", stdin);
100     Input();
101     Solve();
102     return 0;
103 }