题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=44
Dynamic Programming. 对所有可能存在的长方形计算sum，利用已知的小的sub-rectangle计算包含有这些相减或相加得到的new rectangle。代码如下：
1 #include <iostream> 2 #include <math.h> 3 #include <stdio.h> 4 #include <cstdio> 5 #include <algorithm> 6 #include <string.h> 7 #include <string> 8 #include <sstream> 9 #include <cstring> 10 #include <queue> 11 #include <vector> 12 #include <functional> 13 #include <cmath> 14 #include <set> 15 #define SCF(a) scanf("%d", &a) 16 #define IN(a) cin>>a 17 #define FOR(i, a, b) for(int i=a;i<b;i++) 18 #define Infinity 999999999 19 typedef long long Int; 20 using namespace std; 21 22 int main() 23 { 24 int N; 25 int num[105][105]; 26 int sum[105][105]; 27 while (SCF(N) != EOF) 28 { 29 FOR(i, 0, N + 1) 30 num[0][i] = num[i][0] = 0; 31 FOR(i, 1, N + 1) 32 { 33 FOR(j, 1, N + 1) 34 SCF(num[i][j]); 35 } 36 37 int maxSum = sum[0][0]; 38 FOR(i, 0, N + 1) 39 { 40 FOR(j, 0, N + 1) 41 { 42 sum[i][j] = num[i][j]; 43 if (i > 0) 44 sum[i][j] += sum[i - 1][j]; 45 if (j > 0) 46 sum[i][j] += sum[i][j - 1]; 47 if (i > 0 && j > 0) 48 sum[i][j] -= sum[i - 1][j - 1]; 49 maxSum = max(maxSum, sum[i][j]); 50 } 51 } 52 53 FOR(i, 0, (N+1)*(N+1)) 54 { 55 int a = i / (N+1); 56 int b = i % (N + 1); 57 FOR(j, i+1, (N + 1)*(N + 1)) 58 { 59 int c = j / (N + 1); 60 int d = j % (N + 1); 61 if (c >= a && d >= b) 62 { 63 int csum = sum[c][d] - sum[c][b] - sum[a][d] + sum[a][b]; 64 maxSum = max(maxSum, csum); 65 } 66 } 67 } 68 69 printf("%d\n", maxSum); 70 } 71 return 0; 72 }
