UVa 10118 Free Candies（记忆化搜索经典）
Link：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=13&problem=1059&mosmsg=Submission+received+with+ID+15086449
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 Pocket Piles Basket Pocket
1 2 3 4 1 3 4
1 5 6 7 1 5 6 7
2 3 3 3 nothing nothing 2 3 3 3 2 nothing
4 9 8 6 4 9 8 6
8 7 2 1 8 7 2 1
Step3 Take one from pile #2 Step4 Take one from pile #3
Piles Basket Pocket Piles Basket Pocket
1 3 4 1 4
1 6 7 1 6 7
2 3 3 3 2 5 nothing 2 3 3 3 2 3 5 nothing
4 9 8 6 4 9 8 6
8 7 2 1 8 7 2 1
Step5 Take one from pile #2 Step6 Put two candies into his pocket
Piles Basket Pocket Piles Basket Pocket
1 4 1 4
1 6 7 1 6 7
2 3 3 2 3 3 5 nothing 2 3 3 2 5 a pair of 3
4 9 8 6 4 9 8 6
8 7 2 1 8 7 2 1
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 not 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
题意：
有4堆糖果，每堆有n（最多40）个，有一个篮子，最多装5个糖果，我们每次只能从某一堆糖果里拿出一个糖果，
如果篮子里有两个相同的糖果，那么就可以把这两个（一对）糖果放进自己的口袋里，问最多能拿走多少对糖果。糖果种类最多20种.
（黑书 148 免费糖果）
思路：
1. 这一题有点逆向思维的味道，dp[a, b, c, d] 表示从每堆中分别拿 a, b, c, d 个时，最多能拿多少个糖果;
2. 注意一点：当拿到 a, b, c, d 时，不能再拿了，此时结果肯定就会固定。利用这一点性质，采用记忆化搜索能有效的减少重复子结构的计算;
3. 题目是只有 0 0 0 0 这一个出发点的，根据这个出发点进行深搜，最终得出结果。
4. 本题可谓是深搜 + 记忆化搜索的经典，状态不是那么明显，子结构也不是那么好抽象，因为转移的末状态并不是固定的，是在不断的搜索中求出来的;
AC code：
#include <iostream>
#include<cstdio>
#include<cstring>
#define N 41
using
namespace
std;
int
n;
int
dp[N][N][N][N];
int
mat[N][4];
int
DP(
int
*top,
int
st,
int
k)
{
int
&m=dp[top[0]][top[1]][top[2]][top[3]];
if
(m!=-1)
return
m;
if
(top[0]==n&&top[1]==n&&top[2]==n&&top[3]==n||k==5)
return
m=0;
for
(
int
i=0;i<4;i++)
if
(top[i]<n)
{
int
bit=1<<mat[top[i]][i];
top[i]++;
if
(bit&st)
m=max(m,DP(top,st-bit,k-1)+1);
else
if
(k<5)
m=max(m,DP(top,st+bit,k+1));
top[i]--;
}
return
m;
}
int
main()
{
while
(
scanf
(
"%d"
,&n),n)
{
memset
(dp,-1,
sizeof
(dp));
for
(
int
i=0;i<n;i++)
for
(
int
j=0;j<4;j++)
scanf
(
"%d"
,&mat[i][j]);
int
top[5]={0};
printf
(
"%d\n"
,DP(top,0,0));
}
return
0;
}
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