Free Candies
Free Candies
原题链接 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1059
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):
| Step1 | Initial Piles | Step2 | Take one from pile #2 | ||
| Piles | Basket | Piles | Basket | ||
|
nothing | nothing | |
2 | nothing |
| Step3 | Take one from pile #2 | Step4 | Take one from pile #3 | ||
| Piles | Basket | Piles | Basket | ||
|
2 5 | nothing | |
2 3 5 | nothing |
| Step5 | Take one from pile #2 | Step6 | put two candies into his pocket | ||
| Piles | Basket | Piles | Basket | ||
|
2 3 3 5 | nothing | |
2 5 | a pair of 3 |
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?
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.
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
两种做法，可以dp，也可以构建栈来做，直接搜索有重复，时间过大会超限，使用记忆化搜索保存状态，对于每组数据，先初始化vis，d，dp，三个数组，vis数组用来标记，某个糖果如果被放进篮子里便标记为1，四维记忆数组dp代表每堆糖果分别被拿了a，b，c，d个糖果时最多拿走的糖果对数，d数组记录每堆糖果已经被拿走的糖果数，构建完成后最后输出解即可。
以下是AC代码
#include
#include
#include
using namespace std;
int n;
int d[4],vis[100],a[4][100];
int dp[100][100][100][100];
int dfs(int t)
{
if(dp[d[0]][d[1]][d[2]][d[3]]!=-1)
{
return dp[d[0]][d[1]][d[2]][d[3]];
}
if(t==5)
{
return dp[d[0]][d[1]][d[2]][d[3]]=0;
}
int ans=0;
for(int i=0; i<4; i++)
{
if(d[i]==n)
{
continue;
}
int k=a[i][d[i]];
d[i]++;
if(vis[k])
{
vis[k]=0;
ans=max(ans,dfs(t-1)+1);
vis[k]=1;
}
else
{
vis[k]=1;
ans=max(ans,dfs(t+1));
vis[k]=0;
}
d[i]--;
}
return dp[d[0]][d[1]][d[2]][d[3]]=ans;
}
int main()
{
while(~scanf("%d",&n),n)
{
memset(d,0,sizeof d);
memset(dp,-1,sizeof dp);
memset(vis,0,sizeof vis);
for(int i=0; i