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。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 using namespace std;

 5 int dp[2015][2015];

 6 char s1[2011];

 7 int hash[201];

 8 int main()

 9 {

10     int n,m,i,j,a,b;

11     char ch[3];

12     while(scanf("%d%d",&n,&m)!=EOF)

13     {

14         scanf("%s",s1);

15         memset(dp,0,sizeof(dp));

16         for(i = 1;i <= n;i ++)

17         {

18             scanf("%s%d%d",ch,&a,&b);

19             hash[ch[0]-'a'] = min(a,b);

20         }

21         for(i = m;i >= 1;i --)

22         {

23             for(j = i;j <= m;j ++)

24             {

25                 if(s1[i-1] == s1[j-1])

26                 {

27                     dp[i][j] = dp[i+1][j-1];

28                 }

29                 else

30                 {

31                     dp[i][j] = min(dp[i+1][j]+hash[s1[i-1]-'a'],dp[i][j-1]+hash[s1[j-1]-'a']);

32                 }

33             }

34         }

35         printf("%d\n",dp[1][m]);

36     }

37     return 0;

38 }

39 /*

40 2 5

41 bbaab

42 a 1 2

43 b 10 10

44 */