poj3280 Cheapest Palindrome

思路：

区间dp。添加和删除本质相同。

实现：

 #include <iostream>
 #include <cstdio>
 using namespace std;
 int n,m;
 int cost[];
 int dp[][];
 int main()
 {
     string s;
     int x,y;
     char z;
     cin >> n >> m >> s;
     for(int i = ; i < n; i++)
     {
         cin >> z >> x >> y;
         cost[z - 'a'] = min(x, y);
     }
     for(int i = m-; i >= ; i--)
     {
         for(int j = i+; j < m; j++)
         {
             if(s[i] == s[j])
             {
                 dp[i][j] = dp[i+][j-];
             }
             else
             {
                 dp[i][j] = min(dp[i+][j] + cost[s[i] - 'a'],
                                dp[i][j-] + cost[s[j] - 'a']);
             }
         }
     }
     cout << dp[][m-] << endl;
     return ;
 }