【动态规划】bzoj1939: [Croatian2010] Zuma

 1 #include<bits/stdc++.h>
 2 const int maxn = 103;
 3 
 4 int n,m,K,tmp[maxn];
 5 int a[maxn],c[maxn],f[maxn][maxn][maxn];
 6 bool vis[maxn][maxn][maxn];
 7 
 8 int read()
 9 {
10     char ch = getchar();
11     int num = 0;
12     for (; !isdigit(ch); ch=getchar());
13     for (; isdigit(ch); ch=getchar())
14         num = (num<<1)+(num<<3)+ch-48;
15     return num;
16 }
17 inline void Min(int &x, int y){x = x<y?x:y;}
18 int dp(int l, int r, int t)
19 {
20     if (l > r) return 0;
21     if (vis[l][r][t]) return f[l][r][t];
22     vis[l][r][t] = 1;
23     if (a[r]+t >= K) Min(f[l][r][t], dp(l, r-1, 0));
24     else Min(f[l][r][t], dp(l, r, t+1)+1);　
　　　　//这里一开始想成dp(l, r, t-1)+1.实际上这里的状态表示的是逆推。也就是说f[]到完成消除的状态有多远。　
25     for (int k=l; k<r; k++)
26         if (c[k]==c[r])
27             Min(f[l][r][t], dp(l, k, t+1)+dp(k+1, r-1, 0));
28     return f[l][r][t];
29 }
30 int main()
31 {
32     memset(f, 0x3f3f3f, sizeof f);
33     n = read(), K = read();
34     for (int i=1; i<=n; i++)
35         c[i] = read(), a[i] = 1;
36     printf("%d\n",dp(1, n, 0));
37     return 0;
38 }