HDU 5234 背包。

J - 10

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice 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.

 

思路：动态转移方程  dp[i][j][l] = max(dp[i][j-1][l], dp[i-1][j][l], dp[i][j-1][l-a[i][j]]+a[i][j], dp[i-1][j][l-a[i][j]]+a[i][j]);

 

代码：

 

#include<stdio.h>
#include<string.h>
#include<math.h>

#define max(a, b)(a > b ? a : b)
#define N 106
int dp[N][N][N];
int a[N][N];

int main()
{
int i, j, n, m, k, l, aa, b, c, d;

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

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

for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
for(l = 0; l <= k; l++)
{
aa = b = c= d;

aa = dp[i][j-1][l];
b = dp[i-1][j][l];

if(l >= a[i][j])//如果物品的体积小于等于当前背包的体积，。
{
c = dp[i][j-1][l-a[i][j]]+a[i][j];//放入后上边物品的价值。
d = dp[i-1][j][l-a[i][j]]+a[i][j];//放入后左边物品的价值。

dp[i][j][l] = max(max(aa, b),max(c, d));

}
else//如果物品的体积大于当前背包的体积， 就不放， 判断左边的点和上边的点那个大。
dp[i][j][l] = max(aa, b);
}
printf("%d\n", dp[n][m][k]);

}
return 0;
}