uva 10755(dp)

题意：有a*b*c个小长方体块组成一个大长方体块，每个小长方体有一个对应数值，问大长方体的子长方体数值之和最大是多少。

题解：想不出来、、看书上说是枚举上下边界，然后将问题转化成一维问题，但是书上的代码看不懂，然后看了某大神题解后，脑海里大概yy了下立体图形还是比较好理解的。注意s[i][j][k]存了从一个角到i、j、k的总共数值和。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 30;
long long s

;

int main() {
int t;
scanf("%d", &t);
while (t--) {
int a, b, c;
memset(s, 0, sizeof(s));
scanf("%d%d%d", &a, &b, &c);
for (int i = 1; i <= a; i++)
for (int j = 1; j <= b; j++)
for (int k = 1; k <= c; k++) {
scanf("%lld", &s[i][j][k]);
s[i][j][k] += s[i][j][k - 1] + s[i][j - 1][k] + s[i - 1][j][k] + s[i - 1][j - 1][k - 1] - s[i - 1][j][k - 1] - s[i - 1][j - 1][k] - s[i][j - 1][k - 1] ;
}
long long res = -9999999999999;
for (int x1 = 0; x1 < a; x1++)//枚举上下界
for (int x2 = x1 + 1; x2 <= a; x2++)
for (int y1 = 0; y1 < b; y1++)//枚举二维空间的行
for (int y2 = y1 + 1; y2 <= b; y2++) {
long long temp = 0;
for (int z = 1; z <= c; z++) { //枚举列
long long S = (s[x2][y2][z] - s[x1][y2][z]) - (s[x2][y1][z] - s[x1][y1][z]);
res = max(res, S - temp);
temp = min(temp, S);
}
}
printf("%lld\n", res);
if (t)
printf("\n");
}
return 0;
}