[poj]3280 Cheapest Palindrome 题解

Posted 2020-11-16 chiarochinoful

[poj]3280 Cheapest Palindrome

区间dp

题意：

给你长度为m的字符串，其中有n种字符，每种字符都有两个值，分别是插入这个字符的代价，删除这个字符的代价，让你求将原先给出的那串字符变成一个回文串的最小代价。

M<=2000

设 dp[i][j] 为区间 i~j 的回文串的最小代价

现在考虑怎样从别的状态转移到 区间i~j

三种情况

首先 str[i]==str[j] 那么 dp[i][j] = dp[i+1][j-1]

其次 (i+1)~j 是一个回文串 dp[i][j] = dp[i+1][j] + (add[str[i]] / del[str[i]])

最后 i~(j-1) 是一个回文串 dp[i][j] = dp[i][j-1] + (add[str[j]] / del[str[j]])

代码：

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<iostream>
 6 #include<algorithm>
 7 #define APART puts("----------------------")
 8 #define debug 1
 9 #define FILETEST 1
10 #define inf 2010
11 #define ll long long
12 #define ha 998244353
13 #define INF 0x7fffffff
14 #define INF_T 9223372036854775807
15 #define DEBUG printf("%s %d
",__FUNCTION__,__LINE__)
16 
17 namespace chino{
18 
19 inline void setting(){
20 #if FILETEST
21     freopen("_test.in", "r", stdin);
22     freopen("_test.me.out", "w", stdout);
23 #endif
24     return;
25 }
26 
27 inline int read(){
28     char c = getchar(), up = c; int num = 0;
29     for(; c < ‘0‘ || c > ‘9‘; up = c, c = getchar());
30     for(; c >= ‘0‘ && c <= ‘9‘; num = (num << 3) + (num << 1) + (c ^ ‘0‘), c = getchar());
31     return  up == ‘-‘ ? -num : num;
32 }
33 
34 int n, m;
35 char s[inf];
36 int add[inf], del[inf];
37 int dp[inf][inf];
38 
39 inline int main(){
40     n = read(), m = read();
41     scanf("%s", s + 1);
42     for(int i = 1; i <= n; i++){
43         char c = 0;
44         std::cin >> c;
45         add[c * 1] = read();
46         del[c * 1] = read();
47     }
48     int len = strlen(s + 1);
49     for(int i = 2; i <= len; i++){
50         dp[i][i] = 0;
51         for(int j = i - 1; j; j--){
52             dp[j][i] = INF;
53             if(s[i] == s[j])
54                 dp[j][i] = dp[j + 1][i - 1];
55             dp[j][i] = std::min (dp[j][i] , dp[j + 1][i] + std::min (add[s[j] * 1], del[s[j] * 1]));
56             dp[j][i] = std::min (dp[j][i] , dp[j][i - 1] + std::min (add[s[i] * 1], del[s[i] * 1]));
57         }
58     }
59     printf("%d
", dp[1][len]);
60     return 0;
61 }
62 
63 }//namespace chino
64 
65 int main(){return chino::main();}