uva 10755 - Garbage Heap(暴力+优化)

求最大子长方体的价值和。。。

大复杂度水过。。。。

代码如下：

注意： 读入数据的时候请用long long 读入，

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <time.h>
#include <cstdlib>

#define M 25
#define mod 1000000007
#define eps 1e-7
#define INF 0x7fffffff
#define LL long long

using namespace std;

LL st[M][M][M];
int A, B, C;
LL sum(int x1, int x2, int y1, int y2, int z)
{
    LL ans = 0;
    for(int i = x1; i <= x2; ++i)
        for(int j = y1; j <= y2; ++j)
            ans += st[i][j][z];
    return ans;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d%d", &A, &B, &C);
        for(int i = 0; i < A; ++i)
            for(int j = 0; j < B; ++j)
                for(int k = 0; k < C; ++k)
                    scanf("%lld", &st[i][j][k]);
        LL ans = -(1LL<<60);
        for(int x1 = 0; x1 < A; ++x1)
            for(int x2 = x1; x2 < A; ++x2)
                for(int y1 = 0; y1 < B; ++y1)
                    for(int y2 = y1; y2 < B; ++y2)
                    {
                        LL d[M];
                        for(int i = 0; i < C; ++i)
                            d[i] = sum(x1,x2,y1,y2,i);
                        LL maxx = -(1LL<<60), s = 0;
                        for(int i = 0; i < C; ++i)
                        {
                            if(s<0) s = 0;
                            s += d[i];
                            maxx = max(maxx, s);
                        }
                        ans = max(ans, maxx);
                    }
        printf("%lld\n", ans);
        if(t) printf("\n");
    }
    return 0;
}

高效代码（O(n^5)）预处理一下（用空间换时间）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <time.h>
#include <cstdlib>

#define M 25
#define mod 1000000007
#define eps 1e-7
#define INF 0x7fffffff
#define LL long long

using namespace std;

LL st[M][M][M], s[M][M][M];
int A, B, C;
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        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", &st[i][j][k]);

        for(int k = 1; k <= C; ++k)
            for(int i = 1; i <= A; ++i)
                for(int j = 1; j <= B; ++j)
                    s[i][j][k] = st[i][j][k] + s[i-1][j][k] + s[i][j-1][k] - s[i-1][j-1][k];

        LL ans = -(1LL<<60);
        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 d[M];
                        for(int i = 1; i <= C; ++i)
                            d[i] = s[x2][y2][i]-s[x1-1][y2][i]-s[x2][y1-1][i]+s[x1-1][y1-1][i];
                        LL maxx = -(1LL<<60), s = 0;
                        for(int i = 1; i <= C; ++i)
                        {
                            if(s<0) s = 0;
                            s += d[i];
                            maxx = max(maxx, s);
                        }
                        ans = max(ans, maxx);
                    }

        printf("%lld\n", ans);
        if(t) printf("\n");
    }
    return 0;
}