HDU 5234

Description

Today is Gorwin’s birthday. So her mother want to realize her a wish. Gorwin says that she wants to eat many cakes. Thus, her mother takes her to a cake garden. 

The garden is splited into n*m grids. In each grids, there is a cake. The weight of cake in the i-th row j-th column is ${w_{ij}}$ kilos, Gorwin starts from the top-left(1,1) grid of the garden and walk to the bottom-right(n,m) grid. In each step Gorwin can go to right or down, i.e when Gorwin stands in (i,j), then she can go to (i+1,j) or (i,j+1) (However, she can not go out of the garden). 

When Gorwin reachs a grid, she can eat up the cake in that grid or just leave it alone. However she can’t eat part of the cake. But Gorwin’s belly is not very large, so she can eat at most K kilos cake. Now, Gorwin has stood in the top-left grid and look at the map of the garden, she want to find a route which can lead her to eat most cake. But the map is so complicated. So she wants you to help her.

Input

Multiple test cases (about 15), every case gives n, m, K in a single line. 

In the next n lines, the i-th line contains m integers ${w_{i1}},{w_{i{\rm{2}}}},{w_{i3}}, \cdots {w_{im}}$ which describes the weight of cakes in the i-th row 

Please process to the end of file. 

[Technical Specification] 

All inputs are integers. 

1<=n,m,K<=100 

1<=${w_{ij}}$<=100 

Output

For each case, output an integer in an single line indicates the maximum weight of cake Gorwin can eat.

Sample Input

1 1 2
3
2 3 100
1 2 3
4 5 6

Sample Output

0
16


        
 

Hint

 

In the first case, Gorwin can’t eat part of cake, so she can’t eat any cake. In the second case, Gorwin walks though below route (1,1)->(2,1)->(2,2)->(2,3). When she passes a grid, she eats up the cake in that grid. Thus the total amount cake she eats is 1+4+5+6=16.

#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
#define N 112
int a[200][200];
int dp[N][N][N];
int main()
{
    int n,m,kk;


    while(scanf("%d%d%d", &n,&m,&kk) != EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));

        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                scanf("%d",&a[i][j]);

        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                for(int k=0; k<=kk; k++)
                {
                    if(k < a[i][j])
                        dp[i][j][k] = max(dp[i-1][j][k],dp[i][j-1][k]);///当k小于a[i][j]时不拿
                    else
                    {
                        int x = max(dp[i-1][j][k-a[i][j]],dp[i][j-1][k-a[i][j]])+a[i][j];
                        int y = max(dp[i-1][j][k],dp[i][j-1][k]);
                        dp[i][j][k] = max(x,y);
                    }
                }

        printf("%d\n",dp[n][m][kk]);

    }
    return 0;
}///三维数组，因为走得时候是二维的再加上k是三维