[luoguP2890] [USACO07OPEN]便宜的回文Cheapest Palindrome（DP）

传送门

f[i][j] 表示区间 i 到 j 变为回文串所需最小费用

1.s[i] == s[j]　　f[i][j] = f[i + 1][j - 1]

2.s[i] != s[j]　　f[i][j] = min(f[i + 1][j] + cost[s[i]], f[i][j - 1] + cost[s[j]])

开头去掉一个字母和结尾增加一个字母的效果是一样的

所以 cost[s[i]] = min(add[s[i]], del[s[i]])

——代码

 #include <cstdio>
 #include <cstring>
 #include <iostream>

 int n, m;
 int a[], f[][];
 char s[];

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline int min(int x, int y)
 {
     return x < y ? x : y;
 }

 inline int dp(int x, int y)
 {
     if(x >= y) return f[x][y] = ;
     if(f[x][y] ^ -) return f[x][y];
     if(s[x] == s[y]) return f[x][y] = dp(x + , y - );
     else return f[x][y] = min(dp(x + , y) + a[s[x]], dp(x, y - ) + a[s[y]]);
 }

 int main()
 {
     int i;
     char c[];
     n = read();
     m = read();
     scanf("%s", s + );
     for(i = ; i <= n; i++)
     {
         scanf("%s", c);
         a[c[]] = min(read(), read());
     }
     memset(f, -, sizeof(f));
     printf("%d\n", dp(, m));
     return ;
 }