1045. Favorite Color Stripe (30)       
         
 
  
   
    時間限制
   
   
    200 ms
    

   
  
  
   
    記憶體限制
   
   
    65536 kB
    

   
  
  
   
    代碼長度限制
   
   
    16000 B
    

   
  
  
   
    判題程式
   
   
    Standard
   
  
  
   作者
  
  
   CHEN, Yue
   

  
 
 
  
Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

  
It is said that a normal human eye can distinguish about less than 200 different colors, so Eva's favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

  
Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva's favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

  
Input Specification:

  
Each input file contains one test case. For each case, the first line contains a positive integer N (<=200) which is the total number of colors involved (and hence the colors are numbered from 1 to N). Then the next line starts with a positive integer M (<=200) followed by M Eva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (<=10000) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line are separated by a space.

  
Output Specification:

  
For each test case, simply print in a line the maximum length of Eva's favorite stripe.

  Sample Input:
         
6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6
      
        
 Sample Output:        
7      
   

第一個數沒用      

 評測結果       
    

時間	結果	得分	題目	語言	用時(ms)	記憶體(kB)	使用者
8月09日 16:50	答案正确	30	1045	C++ (g++ 4.7.2)	11	436	datrilla

測試點

測試點	結果	用時(ms)	記憶體(kB)	得分/滿分
答案正确	1	300	15/15
1	答案正确	1	308	3/3
2	答案正确	1	308	6/6
3	答案正确	11	436	3/3
4	答案正确	11	436	3/3

#include<iostream> 
#include<vector> 
using namespace std;   
void readln(vector<int>*Which, int Total)
{ 
  for (int index = 0; index < Total; index++)
    cin >> (*Which)[index];
}
void StarCLear(vector<int> *PREmax, int column)
{  
  while (column--)(*PREmax)[column] = 0;
}
void copyNowToPRE(vector<int> *PREmax, int column, vector<int> * Nowmax)
{
  while (column--)(*PREmax)[column] = (*Nowmax)[column];
}
int main()
{  
  int colorTotal,favourTotal,stripeTotal,max;  
  cin >> colorTotal>>favourTotal; 
  vector<int>favorColors(favourTotal);
  readln(&favorColors, favourTotal);
  cin >> stripeTotal;
  vector<int>stripes(stripeTotal);
  readln(&stripes, stripeTotal);
  favourTotal++;
  stripeTotal++;
  vector<int>PREmax(stripeTotal), Nowmax(stripeTotal);
  StarCLear(&PREmax, stripeTotal);
  for (int usecolor = 1; usecolor <favourTotal; usecolor++)/*顔色從i=0~favourable-1；而對應顔色i=usecolor-1最多的放在maxdrix[draw]*/
  {
    for (int draw =1; draw < stripeTotal; draw++)
    {
      max = PREmax[draw]>Nowmax[draw - 1] ? PREmax[draw] : Nowmax[draw - 1];
      Nowmax[draw] = stripes[draw - 1] == favorColors[usecolor - 1] ? max + 1 : max;
    }
    copyNowToPRE(&PREmax, stripeTotal, &Nowmax);
  }
  cout << Nowmax[stripeTotal -1] << endl;;
  system("pause");
  return 0;
}       
   

 評測結果       
    

時間	結果	得分	題目	語言	用時(ms)	記憶體(kB)	使用者
8月09日 16:28	答案正确	30	1045	C++ (g++ 4.7.2)	13	8372	datrilla

測試點

測試點	結果	用時(ms)	記憶體(kB)	得分/滿分
答案正确	1	436	15/15
1	答案正确	1	436	3/3
2	答案正确	1	436	6/6
3	答案正确	12	8372	3/3
4	答案正确	13	8372	3/3

#include<iostream> 
#include<vector> 
using namespace std;   
void readln(vector<int>*Which, int Total)
{ 
  for (int index = 0; index < Total; index++)
    cin >> (*Which)[index];
}
void StarCLear(vector<vector<int>> *maxdrix, int row, int column)
{ 
  while (row--)(*maxdrix)[row][0] = 0;
  while (column--)(*maxdrix)[0][column] = 0;
}
int main()
{  
  int colorTotal,favourTotal,stripeTotal,max;  
  cin >> colorTotal>>favourTotal; 
  vector<int>favorColors(favourTotal);
  readln(&favorColors, favourTotal);
  cin >> stripeTotal;
  vector<int>stripes(stripeTotal);
  readln(&stripes, stripeTotal);
  favourTotal++;
  stripeTotal++;
  vector<vector<int>>maxdrix(favourTotal , vector<int>(stripeTotal ));
  StarCLear(&maxdrix, favourTotal, stripeTotal); 
  for (int usecolor = 1; usecolor <favourTotal; usecolor++)/*顔色從i=0~favourable-1；而對應顔色i=usecolor-1最多的放在maxdrix[usecolor][draw]*/
  {
    for (int draw =1; draw < stripeTotal; draw++)
    {
       max = maxdrix[usecolor-1][draw] > maxdrix[usecolor][draw-1] ? /*【這前一種顔色畫到此處draw】與【下一個顔色畫到這一處前一處】那個更長*/
         maxdrix[usecolor-1][draw] : maxdrix[usecolor][draw-1] ;  
       maxdrix[usecolor][draw]= stripes[draw-1 ] == favorColors[usecolor-1] ?  
        max + 1 : max;
    }
  }
  cout << maxdrix[favourTotal-1][stripeTotal-1] << endl;;
  system("pause");
  return 0;
}