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[30];
int dp[2005][2005];
int main()
{
    string s;
    int x,y;
    char z;
    cin >> n >> m >> s;
    for(int i = 0; i < n; i++)
    {
        cin >> z >> x >> y;
        cost[z - 'a'] = min(x, y);
    }
    for(int i = m-1; i >= 0; i--)
    {
        for(int j = i+1; j < m; j++)
        {
            if(s[i] == s[j])
            {
                dp[i][j] = dp[i+1][j-1];
            }
            else
            {
                dp[i][j] = min(dp[i+1][j] + cost[s[i] - 'a'], 
                               dp[i][j-1] + cost[s[j] - 'a']);
            }
        }
    }
    cout << dp[0][m-1] << endl;
    return 0;    
}