題目連結：https://cn.vjudge.net/problem/UVA-10118

思路：設dp[pa][pb][pc][pd]四堆糖分別取到這四堆的第pa、pb、pc、pd顆時最多的pair數， 用一個二進位制串記錄籃中糖的狀態（即哪些糖有那些沒有）。狀態轉移一共四個，即分別從這四堆中拿出一顆糖放到籃子裡，轉移到放之後的狀態。詳見程式碼

#include<cstdio>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<cstring>
#include<sstream>
#include<iostream>
using namespace std;
const int maxn = 42;

int a[maxn], b[maxn], c[maxn], d[maxn]; //four piles
int DP[maxn][maxn][maxn][maxn];
int n;

inline bool Push(int pre, int m, int &st) // if she can take a pair of candies;
{
    bool ok = false;
    if(pre & (1<<m)) pre ^= (1<<m), ok = true;
    else pre |= (1<<m);
    st = pre;
    return ok;
}

int dp(int pa, int pb, int pc, int pd, int st, int num) // st : sugars that are in the basket
{
    int &ans = DP[pa][pb][pc][pd];
    if(ans >= 0) return ans;
    ans = 0;
    if(num < 5) {
       if(pa < n) {
          int st0;
          if(Push(st, a[pa], st0)) ans = max(ans, 1+dp(pa+1, pb, pc, pd, st0, num-1));
          else ans = max(ans, dp(pa+1, pb, pc, pd, st0, num+1));
       }
       if(pb < n) {
          int st0;
          if(Push(st, b[pb], st0)) ans = max(ans, 1+dp(pa, pb+1, pc, pd, st0, num-1));
          else ans = max(ans, dp(pa, pb+1, pc, pd, st0, num+1));
       }
       if(pc < n) {
          int st0;
          if(Push(st, c[pc], st0)) ans = max(ans, 1+dp(pa, pb, pc+1, pd, st0, num-1));
          else ans = max(ans, dp(pa, pb, pc+1, pd, st0, num+1));
       }
       if(pd < n) {
          int st0;
          if(Push(st, d[pd], st0)) ans = max(ans, 1+dp(pa, pb, pc, pd+1, st0, num-1));
          else ans = max(ans, dp(pa, pb, pc, pd+1, st0, num+1));
       }
    }
    return ans;
}

int main()
{
   while(scanf("%d", &n) && n) {
      memset(DP, -1, sizeof DP);
      for(int i = 0; i < n; i++)
         scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
      printf("%d\n", dp(0, 0, 0, 0, 0, 0));
   }
}