poj3280 Cheapest Palindrome

思路：

区间dp。添加和删除本质相同。

实现：

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int n,m;
 5 int cost[30];
 6 int dp[2005][2005];
 7 int main()
 8 {
 9     string s;
10     int x,y;
11     char z;
12     cin >> n >> m >> s;
13     for(int i = 0; i < n; i++)
14     {
15         cin >> z >> x >> y;
16         cost[z - 'a'] = min(x, y);
17     }
18     for(int i = m-1; i >= 0; i--)
19     {
20         for(int j = i+1; j < m; j++)
21         {
22             if(s[i] == s[j])
23             {
24                 dp[i][j] = dp[i+1][j-1];
25             }
26             else
27             {
28                 dp[i][j] = min(dp[i+1][j] + cost[s[i] - 'a'], 
29                                dp[i][j-1] + cost[s[j] - 'a']);
30             }
31         }
32     }
33     cout << dp[0][m-1] << endl;
34     return 0;    
35 }