Cheapest Palindrome（区间DP）

Link：http://poj.org/problem?id=3280

Cheapest Palindrome

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7180 Accepted: 3468

Description

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.

Input

Line 1: Two space-separated integers: N and M 
Line 2: This line contains exactly M characters which constitute the initial ID string 
Lines 3..N+2: Each line contains three space-separated entities: a character of the input alphabet and two integers which are respectively the cost of adding and deleting that character.

Output

Line 1: A single line with a single integer that is the minimum cost to change the given name tag.

Sample Input

3 4abcba 1000 1100b 350 700c 200 800

Sample Output

900

Hint

If we insert an "a" on the end to get "abcba", the cost would be 1000. If we delete the "a" on the beginning to get "bcb", the cost would be 1100. If we insert "bcb" at the begining of the string, the cost would be 350 + 200 + 350 = 900, which is the minimum.

Source

USACO 2007 Open Gold

题意：给出一个字符串，问通过删除、添加操作使其变为回文串，删除和添加某个字符有特定的代价，求最小代价。

编程思想：

区间DP。dp[i][j]表示使字符串区间i到j为回文串的最小代价，则：

dp1=min(dp[i+1][j]+add[str[i]-'a'],dp[i+1][j]+del[str[i]-'a']);//右边str[j]固定，通过操作左边的str[i]使其成为回文串的最小代价

dp2=min(add[str[j]-'a']+dp[i][j-1],dp[i][j-1]+del[str[j]-'a']);//左边str[i]固定，通过操作右边的str[j]使其成为回文串的最小代价

最后dp[i][j]=min(dp1，dp2);

特别注意当str[i]==str[j]时，除了进行上述操作，还需要再判断：dp[i][j]=min(dp[i][j]，dp[i+1][j-1]);

AC code:

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;int dp[2005][2005];char str[2005];int add[33],del[33];//str[i]!=str[j]:dp[i][j]=min(dp[i+1][j]+add[str[i],dp[i+1][j]+del[str[i]);//str[i]==str[j]:dp[i][j]=dp[i+1][j-1];int main(){   // freopen("D:\\in.txt","r",stdin);    int T,i,j,k,n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        scanf("%s",str+1);        char ch[9];        int tmp1,tmp2;        for(i=1;i<=n;i++)        {            scanf("%s%d%d",ch,&tmp1,&tmp2);            add[ch[0]-'a']=tmp1;            del[ch[0]-'a']=tmp2;        }        memset(dp,0,sizeof(dp));        for(i=m-1;i>=1;i--)//从后往前枚举起点        {            for(j=i+1;j<=m;j++)//枚举终点            {                dp[i][j]=min(dp[i+1][j]+add[str[i]-'a'],del[str[i]-'a']+dp[i+1][j]);//首删除或尾添加                int tmp=min(dp[i][j-1]+del[str[j]-'a'],add[str[j]-'a']+dp[i][j-1]);//尾删除或首添加                dp[i][j]=min(tmp,dp[i][j]);                if(str[i]==str[j])                    dp[i][j]=min(dp[i][j],dp[i+1][j-1]);            }        }        printf("%d\n",dp[1][m]);    }    return 0;}