hdu-5234（dp）

Happy birthday

Problem 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 iswij 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 wi1,wi2,wi3,⋯wim 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<=wij<=100

 

Output

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

 

Sample Input

1 1 232 3 1001 2 34 5 6

 

Sample Output

016

题意：map[n][m]中，每个坐标有一定质量蛋糕，从（1,1）走到（n,m），只可以向左或下走，问吃不超过k质量的蛋糕最多吃多少。每个位置的蛋糕只能不吃或全吃。

思路：

#include<stdio.h>#include<iostream>#include<string.h>#include<math.h>#include<map>#include<vector>#include<iostream>#include<stdio.h>#include<algorithm>using namespace std;int dp[105][105][105];int mp[105][105];int n,m,k;bool cango(int x,int y){    if(x<1||x>n||y<1||y>m) return false;    return true;}int main(){    while(~scanf("%d%d%d",&n,&m,&k))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)                scanf("%d",&mp[i][j]);        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                for(int kk=1;kk<=k;kk++)                {                    if(kk==mp[i][j]) dp[i][j][kk]=kk;                    else{                        if(cango(i-1,j)) dp[i][j][kk]=max(dp[i][j][kk],dp[i-1][j][kk]);                        if(cango(i,j-1)) dp[i][j][kk]=max(dp[i][j][kk],dp[i][j-1][kk]);                        if(kk>mp[i][j]){                                dp[i][j][kk]=max(mp[i][j],dp[i][j][kk]);                            if(cango(i-1,j)) dp[i][j][kk]=max(dp[i][j][kk],dp[i-1][j][kk-mp[i][j]]+mp[i][j]);                            if(cango(i,j-1)) dp[i][j][kk]=max(dp[i][j][kk],dp[i][j-1][kk-mp[i][j]]+mp[i][j]);                        }                    }                }            }        }        printf("%d\n",dp[n][m][k]);    }    return 0;}