ZOJ 1074 To the Max
Time Limit: 2MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
Problem
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.
The input consists of an N x 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.
Example
Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Output
15
思路:
DP求最大和子矩阵
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <string> #include <queue> #include <map> #include <stack> #include <set> #include <cmath> #include <vector> #define INF 0x3f #define eps 1e-6 #define moo 1000000007 using namespace std; int n; int dp[150][150]; int work() { memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { scanf("%d",&dp[i][j]); dp[i][j]+=dp[i][j-1]; } int maxx=0; for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) { int sum=0; for(int k=1;k<=n;k++) { sum+=dp[k][j]-dp[k][i-1]; if(sum>0) { maxx=max(maxx,sum); } else sum=0; } } return maxx; } int main() { while(scanf("%d",&n)!=EOF) { cout<<work()<<endl; } return 0; }