UVA 10755 10755 - Garbage Heap(DP, s)

Farmer John has a heap of garbage formed in a rectangular parallelepiped.
It consists of A×B ×C garbage pieces each of which has a value.
The value of a piece may be 0, if the piece is neither profitable nor
harmful, and may be negative which means that the piece is not just
unprofitable, but even harmful (for environment).
The farmer thinks that he has too much harmful garbage, so
he wants to decrease the heap size, leaving a rectangular nonempty
parallelepiped of smaller size cut of the original heap to maximize the sum of the values of the garbage
pieces in it. You have to find the optimal parallelepiped value. (Actually, if all smaller parallelepiped
has value less than the original one, the farmer will leave the original parallelepiped).
Input
The first line of the input contains the number of the test cases, which is at most 15. The descriptions
of the test cases follow. The first line of a test case description contains three integers A, B, and C
(1 ≤ A, B, C ≤ 20). The next lines contain A · B · C numbers, which are the values of garbage pieces.
Each number does not exceed 231 by absolute value. If we introduce coordinates in the parallelepiped
such that the cell in one corner is (1,1,1) and the cell in the opposite corner is (A, B, C), then the values
are listed in the order
(1, 1, 1),(1, 1, 2), . . . ,(1, 1, C),
(1, 2, 1), . . . ,(1, 2, C), . . . ,(1, B, C),
(2, 1, 1), . . . ,(2, B, C), . . . ,(A, B, C).
The test cases are separated by blank lines.
Output
For each test case in the input, output a single integer denoting the maximal value of the new garbage
heap. Print a blank line between test cases.
Sample Input
1
2 2 2

-1 2 0 -3 -2 -1 1 5

1629140310755Garbage HeapAcceptedC++0.0262015-10-19 08:32:45

ac代码

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;#define ll long long#define INF 0x3f3f3f3f3f3f3f  ll a[22][22][22],sum[22][22][22][22],res[22][22][22][22];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int i,j,k,x,y,z,c,l;        scanf("%d%d%d",&x,&y,&z);        for(i=1;i<=x;i++)            for(j=1;j<=y;j++)                for(k=1;k<=z;k++)                    scanf("%lld",&a[i][j][k]);        ll ans=-INF;        for(c=1;c<=x;c++)            for(i=1;i<=y;i++)                for(j=i;j<=y;j++)                    for(k=1;k<=z;k++)                    {                        ll h=0;                        for(l=k;l<=z;l++)                        {                            h+=a[c][j][l];                            sum[i][j][k][l]=sum[i][j-1][k][l]+h;                            if(c==1)                                res[i][j][k][l]=sum[i][j][k][l];                            else                                res[i][j][k][l]=max(sum[i][j][k][l],res[i][j][k][l]+sum[i][j][k][l]);                            ans=max(ans,res[i][j][k][l]);                        }                    }        printf("%lld\n",ans);        if(t)            printf("\n");    }}