UVA - 108
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.
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 
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.
The input consists of an 
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 
integers separated
by white-space (newlines and spaces). These 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].
The output is the sum of the maximal sub-rectangle.
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
15
Source
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0xffffffff
using namespace std;
int N;
int a[105][105];
int sum[105][105];
int main() {
while(scanf("%d", &N) != EOF) {
for(int i = 1; i <= N; i++) {
for(int j = 1; j <= N; j++) {
scanf("%d", &a[i][j]);
}
}
memset(sum, 0, sizeof(sum));
for(int i = 1; i <= N; i++) {
int t = 0;
for(int j = 1; j <= N; j++) {
t += a[i][j];
sum[i][j] = sum[i - 1][j] + t;
//printf("%d ", sum[i][j]);
}
//printf("%d\n");
}
int p, q, ans = INF;
for(int i = 1; i < N; i++) {
for(int j = 1; j < N; j++) {
for(int k = i; k <= N; k++) {
for(int l = j; l <= N; l++) {
int tmp = sum[k][l] - sum[i-1][l] - sum[k][j-1] + sum[i-1][j-1];
if(tmp > ans) ans = tmp;
}
}
}
}
printf("%d\n", ans);
}
return 0;
}
UVA - 108 - Maximum Sum （简单贪心）
原文地址：http://blog.csdn.net/u014355480/article/details/44838735
