POJ 3280 Cheapest Palindrome

Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow an electronic ID tag that the system will read as the cows pass by a scanner. Each ID tag‘s contents are currently a single string with length M (1 ≤ M ≤ 2,000) characters drawn from an alphabet of N (1 ≤ N ≤ 26) different symbols (namely, the lower-case roman alphabet).

Cows, being the mischievous creatures they are, sometimes try to spoof the system by walking backwards. While a cow whose ID is "abcba" would read the same no matter which direction the she walks, a cow with the ID "abcb" can potentially register as two different IDs ("abcb" and "bcba").

FJ would like to change the cows‘s ID tags so they read the same no matter which direction the cow walks by. For example, "abcb" can be changed by adding "a" at the end to form "abcba" so that the ID is palindromic (reads the same forwards and backwards). Some other ways to change the ID to be palindromic are include adding the three letters "bcb" to the begining to yield the ID "bcbabcb" or removing the letter "a" to yield the ID "bcb". One can add or remove characters at any location in the string yielding a string longer or shorter than the original string.

Unfortunately as the ID tags are electronic, each character insertion or deletion has a cost (0 ≤ cost ≤ 10,000) which varies depending on exactly which character value to be added or deleted. Given the content of a cow‘s ID tag and the cost of inserting or deleting each of the alphabet‘s characters, find the minimum cost to change the ID tag so it satisfies FJ‘s requirements. An empty ID tag is considered to satisfy the requirements of reading the same forward and backward. Only letters with associated costs can be added to a string.

区间动态规划

我们可以设dp[i][j]代表将I到J这一段变为回文的最小代价，然后便可以发现它可以由以下三种情况转移来得到：

1.若a[l]==a[r]   可以直接由dp[l+1][r-1]转移而来（因为dp[l+1][r-1]已经在前面更新过，已经是回文的了）.

2.若a[l+1]==a[r] 那么我们可以通过删除a[l]或者通过添加一个与a[i]相同的字符来完成对dp[l][r]的修改，使其成为回文串，即dp[l][r]=min(dp[l][r],dp[l+1][r]+str[a[l]]).

3.若 a[l]==a[r-1]  与2同理，稍稍做些修改即可.

   转移时注意要由先枚举长度，由小区间取更新大区间来得到大区间的最优值。最后的答案就是dp[0][m-1].

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#define RG register
#define IL inline
#define pi acos(-1.0)
#define ll long long 
using namespace std;
int n,m;
char a[2500];
map <char,int> s;
int dp[2500][2500];//代表将I到J这一段变为回文的最小代价
int main() {
  scanf("%d%d",&n,&m);
  scanf("%s",a);
  for(int i=1;i<=n;i++){
     char str;
     int del,add;
     cin>>str;
     scanf("%d%d",&add,&del);
     s[str]=min(add,del);
  }
  memset(dp,50,sizeof(dp));
  for(int i=0;i<m;i++)  dp[i][i]=0;
  for(int i=2;i<=m;i++)//拿已经是回文的小区间取更新大的区间
    for(int l=0,r=0+i-1;r<m;++l,++r){
      if(a[l]==a[r]){
    if((l+1)>(r-1)) dp[l][r]=0;
    else dp[l][r]=dp[l+1][r-1];
    //已经是回文了，直接赋值即可
      } else{
    dp[l][r]=min(dp[l][r],dp[l+1][r]+s[a[l]]);
    dp[l][r]=min(dp[l][r],dp[l][r-1]+s[a[r]]);
        //把原来不是回文的变成回文的
      }
    }
  printf("%d",dp[0][m-1]);
  return 0;
}