HDU 5900 - QSC and Master [ DP ]

Posted 2020-08-10 nicetomeetu

题意：   

　　给n件物品，有key和value   

　　每次可以把相邻的 GCD(key[i], key[i+1]) != 1 的两件物品，问移除的物品的总value最多是多少   

　　key : 1 3 4 2  移除34以后12也相邻了，可以移除

分析：   

　　先预处理出所有GCD[l][r]， 意味 l <= i <= r的区域可以全部移除， 用记忆化搜索处理   

　　然后 dp[i] 代表到 i 为止可以得到的最大value和   

　　if (G[l][r]) dp[r] = max(dp[r], dp[l-1] + sum[r] - sum[l-1] )

　　以及用 dp[i] = max(dp[i], dp[i-1]) 向后保留最大值

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 #define LL long long
 6 const int MAXN = 305;
 7 int t, n;
 8 LL dp[MAXN];
 9 LL key[MAXN], v[MAXN], sum[MAXN];
10 int GCD[MAXN][MAXN];
11 LL gcd(LL a, LL b)
12 {
13     return b == 0 ? a : gcd(b, a%b); 
14 }
15 bool check(int l, int r)
16 {
17     if (l > r) return 1;
18     if (GCD[l][r] != -1) return GCD[l][r];
19     if (l == r) return GCD[l][r] = 0;
20     if ((r-l)%2 == 0) return GCD[l][r] = 0; 
21     if (l == r-1) return GCD[l][r] = ( gcd(key[l], key[r]) != 1 );
22     if (gcd(key[l], key[r]) != 1 && check(l+1, r-1) ) return GCD[l][r] = 1;
23     for (int i = l+1; i < r-1; i++)
24     {
25         if (check(l,i) && check(i+1, r)) return GCD[l][r] = 1;
26     }
27     return GCD[l][r] = 0;
28 }
29 int main()
30 {
31     scanf("%d", &t);
32     while (t--)
33     {
34         scanf("%d", &n);
35         for (int i = 1; i <= n; i++)
36             scanf("%lld", &key[i]);
37         sum[0] = 0;
38         for (int i = 1; i <= n; i++)
39             scanf("%lld", &v[i]) , sum[i] = sum[i-1] + v[i];
40         memset(GCD, -1, sizeof(GCD));
41         for (int i = 1; i <= n; i++)
42             for (int j = i; j <= n; j++)
43                 check(i, j);
44         memset(dp, 0, sizeof(dp));
45         for (int i = 1; i <= n; i++)
46         {
47             for (int j = 2; j <= i; j += 2)
48             {
49                 int l = i - j + 1, r = i;
50                 if (check(l, r))
51                     dp[i] = max(dp[i], dp[l-1] + sum[r] - sum[l-1]);
52             }
53             dp[i] = max(dp[i], dp[i-1]);
54         }
55         printf("%lld\n", dp[n]);
56     }
57 }