uva 10118 Free Candies

Problem Description

Little Bob is playing a game. He wants to win some candies in it - as

many as possible.

There are 4 piles, each pile contains N candies. Bob is given a basket

which can hold at most 5 candies. Each time, he puts a candy at the

top of one pile into the basket, and if there’re two candies of the

same color in it ,he can take both of them outside the basket and put

them into his own pocket. When the basket is full and there are no two

candies of the same color, the game ends. If the game is played

perfectly, the game will end with no candies left in the piles.

The Input

The input will contain no more than 10 test cases. Each test case

begins with a line containing a single integer n(1<=n<=40)

representing the height of the piles. In the following n lines, each

line contains four integers xi1,xi2,xi3,xi4 (in the range 1..20). Each

integer indicates the color of the corresponding candy. The test case

containing n=0 will terminate the input, you should not give an answer

to this case.

The Output

Output the number of pairs of candies that the cleverest

little child can take home. Print your answer in a single line for

each test case.

Sample Input

5

1 2 3 4

1 5 6 7

2 3 3 3

4 9 8 6

8 7 2 1

1

1 2 3 4

3

1 2 3 4

5 6 7 8

1 2 3 4

0

Sample Output

8

0

3

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 50;
int n;
int f[maxn][maxn][maxn][maxn];
int piles[maxn][maxn];

int dp(int* r,int s,int* t)
{
if(s==5) return f[t[0]][t[1]][t[2]][t[3]]=0;
if(f[t[0]][t[1]][t[2]][t[3]] >= 0)return f[t[0]][t[1]][t[2]][t[3]];
f[t[0]][t[1]][t[2]][t[3]] = 0;
for(int i = 0; i < 4 ; i++ )
{
if(t[i]==n)continue;
if(r[piles[i][t[i]]]==1)
{
int &v = f[t[0]][t[1]][t[2]][t[3]];
r[piles[i][t[i]]]=0;
t[i]++;
v = max(v,dp(r,s-1,t) + 1);
t[i]--;
r[piles[i][t[i]]]=1;
}
else
{
int& v = f[t[0]][t[1]][t[2]][t[3]];
r[piles[i][t[i]]]=1;
t[i]++;
v = max(v,dp(r,s+1,t));
t[i]--;
r[piles[i][t[i]]]=0;
}
}
return f[t[0]][t[1]][t[2]][t[3]];
}

int main()
{
while((cin >> n)&&n)
{
memset(f,-1,sizeof(f));
for(int i = 0 ; i < n ; i++ )
for(int j = 0 ; j < 4 ; j++ )
cin >> piles[j][i];
int r[25],top[4]={0,0,0,0};
memset(r,0,sizeof(r));
int ans = dp(r,0,top);
cout << ans << endl;
}
return 0;
}