2013
12-12

Free Candies

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.

For example, Bob may play this game like this (N=5):

Note that different numbers indicate different colors, there are 20 kinds of colors numbered 1..20.

‘Seems so hard…’Bob got very much puzzled. How many pairs of candies could he take home at most?

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.

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.

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

8
0
3

/*

http://acm.hdu.edu.cn/showproblem.php?pid=1514

dp[x[1]][x[2]][x[3]][x[4]]记录4堆分别从x[1],x[2],x[3],x[4]处往下取所获得的最大值；

*/

#include<stdio.h>
#include<string.h>
int num[45][5],n;
int x[5],dp[45][45][45][45];
int dfs(int market,int cnt)
{
int sum,i,bit,Max=0;
if(dp[x[1]][x[2]][x[3]][x[4]]!=-1)
return dp[x[1]][x[2]][x[3]][x[4]];
for(i=1;i<=4;i++)
{
sum=0;
x[i]++;
if(x[i]<=n){
bit=(1<<num[x[i]][i]);
if(market&bit)
sum=dfs(market&(~bit),cnt-1)+1;
else if(cnt<4)
sum=dfs(market|bit,cnt+1);
}
if(sum>Max)Max=sum;
x[i]--;
}
dp[x[1]][x[2]][x[3]][x[4]]=Max;
return Max;
}
int main()
{
int i,j;
while(scanf("%d",&n),n)
{
for(i=1;i<=n;i++)
for(j=1;j<=4;j++)
scanf("%d",&num[i][j]);
x[1]=x[2]=x[3]=x[4]=0;
memset(dp,-1,sizeof(dp));
printf("%d\n",dfs(0,0));
}
return 0;
}

1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。