UVa 10755 Garbage Heap (暴力+前缀和)

题意：有个长方体由A*B*C组成，每个废料都有一个价值，要选一个子长方体，使得价值最大。

析：我们暴力枚举上下左右边界，然后用前缀和来快速得到另一个，然后就能得到长方体，每次维护一个最小值，然后差就是最大值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1LL << 60;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 20 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int A, B, C;
LL sum[maxn][maxn][maxn];

void solve(int i){
    A = i & 1;  i >>= 1;
    B = i & 1;  i >>= 1;
    C = i & 1;
}

inline int sign(){  return (A + B + C) & 1 ? 1 : -1;  }

LL getSum(int x1, int x2, int y1, int y2, int z1, int z2){
    int dx = x2 - x1 + 1;
    int dy = y2 - y1 + 1;
    int dz = z2 - z1 + 1;

    LL ans = 0;
    for(int i = 0; i < 8; ++i){
        solve(i);
        ans -= sum[x2-A*dx][y2-B*dy][z2-C*dz] * sign();
    }
    return ans;
}

int main(){
    int T;  cin >> T;
    while(T--){
        int a, b, c;
        scanf("%d %d %d", &a, &b, &c);
        memset(sum, 0, sizeof sum);
        for(int i = 1; i <= a; ++i)  for(int j = 1; j <= b; ++j)
            for(int k = 1; k <= c; ++k)
                scanf("%lld", &sum[i][j][k]);

        for(int i = 1; i <= a; ++i)  for(int j = 1; j <= b; ++j)
            for(int k = 1; k <= c; ++k)
                for(int x = 1; x < 8; ++x){
                    solve(x);
                    sum[i][j][k] += sum[i-A][j-B][k-C] * sign();
                }

        LL ans = -LNF;
        for(int x1 = 1; x1 <= a; ++x1)  for(int x2 = x1; x2 <= a; ++x2)
            for(int y1 = 1; y1 <= b; ++y1)  for(int y2 = y1; y2 <= b; ++y2){
                LL mmin = 0;
                for(int z = 1; z <= c; ++z){
                    LL s = getSum(x1, x2, y1, y2, 1, z);
                    ans = max(ans, s - mmin);
                    mmin = min(mmin, s);
                }
            }

        printf("%lld\n", ans);
        if(T)  putchar('\n');
    }
    return 0;
}