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Language: Default Cheapest Palindrome
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 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 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). 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 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 4 abcb a 1000 1100 b 350 700 c 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 |
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 2005
int dp[N][N],n,m;
int add[N],del[N];
char a[N],ch[N];
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
scanf("%s",a);
int x,y;
for(i=0;i<n;i++)
{
scanf("%s%d%d",ch,&x,&y);
add[ch[0]-'a']=x;
del[ch[0]-'a']=y;
}
memset(dp,0,sizeof(dp));
for(i=m-1;i>=0;i--)
for(j=i+1;j<m;j++)
{
dp[i][j]=min(dp[i+1][j]+del[a[i]-'a'],dp[i+1][j]+add[a[i]-'a']);
int temp=min(dp[i][j-1]+del[a[j]-'a'],dp[i][j-1]+add[a[j]-'a']);
dp[i][j]=min(temp,dp[i][j]);
if(a[i]==a[j])
dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
}
printf("%d\n",dp[0][m-1]);
}
return 0;
}