POJ 3280 Cheapest Palindrome(区间DP)

Cheapest Palindrome

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6861 Accepted: 3327

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

经典的区间DP，对于每个字符，在原字符串加上这个字符的代价是一个值，移除又是一个值，求把原字符串变成回文串的最小代价。经典的区间DP，状态转移方程见代码。在输入进行了一个处理，我们把对一个字符的增与删的操作的代价压缩成为一个数，代表对该字符进行增或删代价，把另一个相对较大的代价则忽略掉。因为在一遍插入一个字符与在另一边删除一个同样的字符对回文串形成的贡献效果一样（可以仔细思考一下）。

#include<stack>#include<queue>#include<cmath>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#pragma commment(linker,"/STACK: 102400000 102400000")#define lson a,b,l,mid,cur<<1#define rson a,b,mid+1,r,cur<<1|1using namespace std;typedef long long LL;const double eps=1e-6;const int MAXN=2100;char s[MAXN];int n,m,c[26],dp[MAXN][MAXN];int main(){    scanf("%d%d%s",&m,&n,s+1);    for(int i=0;i<m;i++)    {        int t1,t2;        char t[3];        scanf("%s%d%d",t,&t1,&t2);        c[t[0]-'a']=min(t1,t2);//忽略较大的代价    }    memset(dp,0,sizeof(dp));    for(int j=2;j<=n;j++)        for(int i=j-1;i>=1;i--)        {            if(s[i]==s[j])//状态转移方程                dp[i][j]=dp[i+1][j-1];            else                dp[i][j]=min(dp[i+1][j]+c[s[i]-'a'],dp[i][j-1]+c[s[j]-'a']);        }    printf("%d\n",dp[1][n]);    return 0;}