本文已参与「新人创作礼」活动，一起开启掘金创作之路。
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 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.
数据：
Sample Input
3 4 abcb a 1000 1100 b 350 700 c 200 800Sample Output
900
输入n和m 然后输入m（<=2000）长度的str字符串 接下来n行 字符c 整型a 整型 b（意思是字符c增加或删除的代价）
题意：通过字符的变化，最后变成回文串，且花费代价最小 方法：区间dp的应用
题解：1：删除和增加在本质上相同的，所以在记录成本是，只要取其中最小的即可。
2：状态转移方程如下，str为原串数组，cost数组为花费的代价，dp[i][j]表示从i字符~j字符构成回文的最小代价
if(str[i]==str[j])
dp[j][i]=dp[j+1][i-1];
else dp[j][i]=min(dp[j+1][i]+cost[str[j]],dp[j][i-1]+cost[str[i]]);
3： 最后在更新dp数组的时候要注意顺序，使子状态都是已经更新过的状态！！
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
const int N=1e5+10;
char str[2020];
int cost[150];///防止数组越界 因为是字符
using namespace std;
int dp[2020][2020];
int n,m;
int solve()
{
mem(dp,0);
for(int i=1; i<m; i++)
{
for(int j=i-1; j>=0; j--)
{
if(str[i]==str[j])
dp[j][i]=dp[j+1][i-1];
else
dp[j][i]=min(dp[j+1][i]+cost[str[j]],dp[j][i-1]+cost[str[i]]);
}
}
return dp[0][m-1];
}
int main()
{
scanf("%d %d",&n,&m);
scanf("%s",str);
char c;
int a,b;
getchar();///！！
for(int i=0; i<n; i++)
{
scanf("%c %d %d",&c,&a,&b);
getchar();///！！
cost[c]=min(a,b);
}
printf("%d\n",solve());
return 0;
}
