POJ 3280 Cheapest Palindrome(DP)

题目链接

被以前的题目惯性思维了，此题dp[i][j]，代表i到j这一段变成回文的最小花费。我觉得挺难的理解的。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 using namespace std;
 int dp[][];
 char s1[];
 int hash[];
 int main()
 {
     int n,m,i,j,a,b;
     char ch[];
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         scanf("%s",s1);
         memset(dp,,sizeof(dp));
         for(i = ;i <= n;i ++)
         {
             scanf("%s%d%d",ch,&a,&b);
             hash[ch[]-'a'] = min(a,b);
         }
         for(i = m;i >= ;i --)
         {
             for(j = i;j <= m;j ++)
             {
                 if(s1[i-] == s1[j-])
                 {
                     dp[i][j] = dp[i+][j-];
                 }
                 else
                 {
                     dp[i][j] = min(dp[i+][j]+hash[s1[i-]-'a'],dp[i][j-]+hash[s1[j-]-'a']);
                 }
             }
         }
         printf("%d\n",dp[][m]);
     }
     return ;
 }
 /*
 2 5
 bbaab
 a 1 2
 b 10 10
 */