poj2955 区间dp

 1 //Accepted    200 KB    63 ms
 2 //区间dp
 3 //dp[i][j] 从i位到j位能得到的最大匹配数
 4 //dp[i][j]=max(dp[i+1][j-1] (s[i-1]==s[j-1]),dp[i][k]+dp[k+1][j])i<=k<j
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <iostream>
 8 using namespace std;
 9 const int imax_n = 105;
10 int dp[imax_n][imax_n];
11 char s[imax_n];
12 int n;
13 int max(int a,int b)
14 {
15     return a>b?a:b;
16 }
17 int isMatch(char ch1,char ch2)
18 {
19     if (ch1==‘(‘ && ch2==‘)‘) return 1;
20     if (ch1==‘[‘ && ch2==‘]‘) return 1;
21     return 0;
22 }
23 void Dp()
24 {
25     for (int i=1;i<=n;i++)
26     dp[i][i]=0;
27     for (int l=2;l<=n;l++)
28     {
29         for (int i=1;i<=n;i++)
30         {
31             int j=i+l-1;
32             if (j>n) break;
33             dp[i][j]=0;
34             if (isMatch(s[i-1],s[j-1])) dp[i][j]=dp[i+1][j-1]+2;
35             for (int k=i;k<j;k++)
36             dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
37         }
38     }
39     printf("%d\n",dp[1][n]);
40 }
41 int main()
42 {
43     while (scanf("%s",s) && s[0]!=‘e‘)
44     {
45         n=strlen(s);
46         Dp();
47     }
48     return 0;
49 }