poj 3280 回文字符串问题 dp算法

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
using namespace std;
int dp[2048][2048];
int cost[48];
int main()
{
        int n, m;
        cin >> n >> m;
        string s;
        cin >> s;
        for (int i = 0; i < n; i++)
        {
               int add, del;
               char c;
               cin >> c >> add >> del;
               cost[c - ‘a‘] = min(add, del);
        }
        for(int i=m-1;i>=0;--i)
               for (int j = i + 1; j < m; j++)
               {
                       dp[i][j] = min(dp[i + 1][j] + cost[s[i] - ‘a‘], dp[i][j - 1] + cost[s[j] - ‘a‘]);
                       if (s[i] == s[j])
                              dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);
               }
        printf("%d\n", dp[0][m - 1]);
        return 0;
}