POJ3280_Cheapest Palindrome_DP

Cheapest Palindrome

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9538 Accepted: 4588

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.

/*************************************************给出一个字符串，通过增加或删除给出的字符，把它变成回文串。其中，每一个字符的增加、删除操作有一定花费要求总的花费最少dp[i][j] 保存的是把区间[i,j]变成回文串的最小花费枚举i和j，长度从小到大填写所有区间状态转移if(str[i] == str[j]) dp[j][i] = dp[j+1][i-1];else dp[j][i] = min(dp[j][i-1] + cost[ str[i]-'a' ], dp[j+1][i] + cost[ str[j]-'a' ]);如果j,i两端元素相同，dp[j,i]=dp[j+1][i-1]因为在这之前dp[j+1][i-1]已经填过，所以可以认为j+1,i-1已经是一个回文串。[j,i]就是在它的两端加上一对相同的元素，不需要增加花费如果不相同与上同理可以认为 [j+1,i] 和 [j][i-1] 已经是回文串。要把[j,i]变成回文串，可以在左边删去j元素变成 [j+1,i], 或者在右端加上一个j元素也行。变成[j][i-1]也同理。只要在两者中选择较小花费即可*************************************************/#include<cstdio>#include<iostream>#include<cstring>using namespace std;int dp [2010][2010];//使区间ij变为回文串的最小花费int cost [30];//增加或删除一个字符的最小花费int N, M;//原串长度，给定字符数char str [2010];//原串int main (){scanf("%d %d", &M, &N);scanf(" %s", str);int L = strlen(str);for(int i= 0; i< M; i++){char s;int x, y;scanf(" %c %d %d", &s, &x, &y);//增加和删除一个字符串操作是等价的//选择其中花费较小的那种cost[ s-'a' ] = min(x, y);}//枚举0到n的每一个区间，求最小花费for(int i= 1; i< L; i++){for(int j= i-1; j>= 0; j--){//区间两端相同if(str[i] == str[j]) dp[j][i] = dp[j+1][i-1];//区间两端不同，改变首元素或尾元素else dp[j][i] = min(dp[j][i-1] + cost[ str[i]-'a' ], dp[j+1][i] + cost[ str[j]-'a' ]);}}printf("%d\n", dp[0][L-1]);return 0;}