POJ 3280 Cheapest Palindrome(水题)

题意：给定一个完全由小写字母组成的字符串s，对每个字母比如x(或a,b,c...z)，在字符串中添加或者删除它分别需要花费c1['x']和c2['x']的代价，问将给定字符串变成回文串所需要的最少代价为多少。

解法：设d[i][j]表示将字符串中从第i位至第j位变成回文串所需要的代价。若s[i] == s[j]，d[i][j] = d[i+1][j-1]；否则的话，有四种处理方法。

　　　对xa.......by，可以将其变为xa......b，yxa.....by，a.......by，xa.....byx中的任意一种再处理。所以d[i][j] = min(d[i][j-1] + c2[s[j]], d[i][j-1] + c1[s[j]], d[i+1][j] + c2[s[i]], d[i+1][j] + c1[s[i]])。化简为d[i][j] = min(d[i][j-1] + min(c1[s[j]], c2[s[j]]), d[i+1][j] + min(c1[s[i]], c2[s[i]]))。(此公式看着没有化简，但实际上可以省很多代码，我的代码改了以后少了近200b)

tag:字符串，回文串，dp

 /*
  * Author:  Plumrain
  * Created Time:  2013-11-17 21:44
  * File Name: DP-POJ-3280.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <map>
 #include <utility>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 const int maxint = ;

 char s[];
 int d[][];
 map<char, int > mp;

 int main()
 {
     int n, len;
     while (scanf ("%d%d", &n, &len) != EOF){
         scanf ("%s", s);
         char x;
         int t1, t2;
         mp.clear();
         for (int i = ; i < n; ++ i){
             x = 'P';
             while (!(x >= 'a' && x <= 'z')) scanf ("%c", &x);
             scanf ("%d%d", &t1, &t2);
             mp[x] = min(t1, t2);
         }

         CLR (d);
         for (int i = ; i < len; ++ i)
             d[i][i] = ;
         for (int i = len-; i >= ; -- i)
             for (int j = i+; j < len; ++ j){
                 if (s[i] == s[j]) d[i][j] = d[i+][j-];
                 else
                     d[i][j] = min(maxint, min(d[i+][j] + mp[s[i]], d[i][j-] + mp[s[j]]));
             }
         printf ("%d\n", d[][len-]);
     }
     return ;
 }