每次看书都是看一遍 ，自己总是眼高手低，以后每次都要自己独自写一下，在做题中提高自己的代码能力。
总结以前，自己不认真思考，想一会就看题解，这样提高不了，自己一定要有充分的思考后动手尝试把自己的思路转化为代码，试一试。别慌，也可以拿纸画一画，写一写，总之以后自己要留足思考的时间。
Cheapest Palindrome
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 18140 Accepted: 8451
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
思路：虽然是书上的例题，上毛概的时候看的，然后晚上自己找到题做一下，确实理解的更深了一点，我们定义dp[i][j]表示从第i个位置到第j个位置花费的最小费用，然后向外扩张，
分为三种情况
- 若str[i]==str[j],则 dp[i][j]=dp[i+1][j-1]；
- 若dp[i+1][j]符合条件，则dp[i][j]= dp[i+1][j]+val[str[i]-‘0’]；
- 若dp[i][j-1]符合条件，则dp[i][j]=dp[i][j-1]+val[str[j]-‘0’];
此外，我们还要初始化一下dp[0][0]=1；
详细看代码：
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
int n,m;
int dp[2005][2005];
int val[1000];
string str;
int main()
{
char s;int x,y;cin>>m>>n;cin>>str;while(m--){
cin>>s>>x>>y;val[s-'0']=min(x,y);}memset(dp,-1,sizeof(dp));dp[0][0]=1;for(int i=n-1;i>=0;i--)for(int j=i;j<n;j++){
if(str[i]==str[j])dp[i][j]=dp[i+1][j-1];elsedp[i][j]=min(dp[i+1][j]+val[str[i]-'0'],dp[i][j-1]+val[str[j]-'0']);}cout<<dp[0][n-1]+1;return 0;
}