dp问题：采药

 

今天把采药A了，属于dp问题，主要思路是把所有时间都存进一个数组中，数组的值对应药的价值，下标是时间，然后记忆化搜索，碰到价值高的就赋值，相比于摘花生，辰辰是一个聪明的猴子，不废话了，上代码。

Code：

#include <iostream>

#include <cstring>

using namespace std;

int ji[1002];

int main()  {

         int t,m,a,b,i,j;

         while(cin>>t>>m)

    {

                   memset(ji,0,sizeof(ji));

                   for(i=1;i<=m;i++)

                   {

                            cin>>a>>b;

                            if(t-a>=0)

            for(j=t-a;j>=0;j--)

             if(ji[j]+b>ji[j+a])

               ji[j+a]=ji[j]+b;

        }

                   cout<<ji[t]<<endl;

         }

         return 0;

 }

附带另一个问题，多多摘花生，这里不吐槽那猴子有多笨了，下面的数据足以说明；

2   8  8

6  0  0  0  0  0  0  0

0         0  0  0  0  0  0  8

那只猴子如果聪明的话能摘14个，笨蛋猴子只能摘8个。

Code：

#include <iostream>

#include <cstdlib>

using namespace std;

int main()

{

        int n;

        int row, col;

        int time;

        cin >> n;

        while (n--)

        {

               cin >> row >> col >> time;

               int i, j;

               int curPi = 0, curPj;

               int p[51][51];

               int timet = 0;

               int pg = 0;

               for (i = 1; i <= row; i++)

               {

                       for (j = 1; j <= col; j++)

                       {

                               cin >> p[j];

                       }

               }

               int maxPi = 0, maxPj = 0;

               while (timet <= time)

               {

 

                       int max = 0;

                       for (i = 1; i <= row; i++)

                       {

                               for (j = 1; j <= col; j++)

                               {

                                      if (p[j] > max)

                                      {

                                              max = p[j];

                                              maxPi = i;

                                              maxPj = j;

                                      }

                               }

 

                       }

                       if (max == 0)

                               break;

                       if (curPi == 0)

                               curPj = maxPj;

                       if (timet +

                               (abs(maxPj - curPj) + abs(maxPi - curPi) + 1 + maxPi) <= time)

                       {

                               timet = timet + abs(maxPj - curPj) + abs(maxPi - curPi) + 1;

                               curPi = maxPi;

                               curPj = maxPj;

                               pg += p[curPi][curPj];

                               p[curPi][curPj] = 0;

                       }

                       else

                               break;

               }

               cout << pg << endl;

        }

        return 0;

 

}

代码还老长。。。。。