POJ 1050 To the Max
To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 35476 | Accepted: 18625 |
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.
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
代码:
// 最大连续和扩展 // 最近DP D到吐血= = 对于新手来说好难想到状态 // 还是好好修炼吧= = #include<cstdio> #include<cstring> using namespace std ; #define M 110 int a[M][M] , d[M] ; int n ; int Sum() { int max = -3724 , i ; int mm = 0 ; for( i = 1; i <= n ;i++) { mm += d[i] ; if( mm > max ) max = mm ; if( mm < 0 ) mm = 0; } return max ; } int dfs() { int i , j , ans = -1213 , cmp ; for( i = 1; i <= n ;i++) { memset( d , 0 , sizeof(d) ) ; for( j = i ; j <= n ;j++)// 枚举行的所以情况 { for( int v = 1; v <= n ; v++) d[v] += a[j][v] ; cmp = Sum() ; // 列用 最大连续和处理 //printf( "%d\n" , cmp ) ; if( ans < cmp ) ans = cmp ; } } return ans ; } int main() { int i , j ; while( scanf( "%d" , &n) != EOF ) { for( i = 1; i <= n ;i++) for( j = 1 ;j <= n ;j++) scanf( "%d" , &a[i][j] ) ; printf( "%d\n" , dfs( ) ) ; } }