POJ 3280 Cheapest Palindrome (DP), pojpalindrome

POJ 3280 Cheapest Palindrome (DP), pojpalindrome
Zookeeper

Description

Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automatic it. he has installed on each cow an electronic ID tag that the system will read as the cows pass by a packet. each ID tag's contents are currently a single string with lengthM(1 ≤M≤ 2,000) characters drawn from an alphabetN(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" wocould 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 wowould like to change the cows's ID tags so they read the same no matter which direction the cow walks. 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 ining 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: NAnd M
Line 2: This line contains exactly MCharacters 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

Question: a string of letters. After adding or deleting a character, the sequence becomes a back-to-text. adding or deleting a string will cost you at least.

Set dp [I] [j] As the cost from I to j,

dp[i][j] = min ( dp[i+1][j]+cost[i] , dp[i][j-1]+cost[j] );  ( a[i] != a[j] )

dp[i][j] = dp[i+1][j-1] ( a[i] == a[j] )

In cost [], each character is deleted or a smaller value is added, because the effect of deleting a [I] is the same as that of adding a [I] After j, you only need to compare the cost of the two.

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;typedef long long LL;const int MAX=0x3f3f3f3f;int n,m,cost[30],dp[2007][2005];char s[2005],cc[3];int main(){    scanf("%d%d%s",&n,&m,s);    for(int i=0;i<n;i++) {        int xx,yy;        scanf("%s %d %d",cc,&xx,&yy);        cost[ cc[0]-'a' ] = min(xx,yy);    }    for(int j=1;j<m;j++)        for(int i=j-1;i>=0;i--)             if( s[i] == s[j] ) dp[i][j] = dp[i+1][j-1];             else dp[i][j] = min( dp[i+1][j]+cost[ s[i]-'a' ] ,dp[i][j-1]+cost[ s[j]-'a' ] );    printf("%d\n",dp[0][m-1]);    return 0;}