poj 3280Cheapest Palindrome

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.

题意：给出一个由m中字母组成的长度为n的串，给出m种字母添加和删除花费的代价，求让给出的串变成回文串的代价。

题解：

我们知道求添加最少的字母让其回文是经典dp问题，转化成LCS求解。这个是一个很明显的区间dp
我们定义dp [ i ] [ j ] 为区间 i 到 j 变成回文的最小代价。
那么对于dp【i】【j】有三种情况
首先：对于一个串如果s【i】==s【j】，那么dp【i】【j】=dp【i+1】【j-1】
其次：如果dp【i+1】【j】是回文串，那么dp【i】【j】=dp【i+1】【j】+min（add【i】，del【i】）；

在一篇博客上看看到了这个：

因为dp[i][j]需要借用i+1时的结果和j-1时的结果。

博主用的是 i需要反向枚举，j需要正向枚举 循环实现的，

看的其他的博主写的（代码2）也可以，自己可以看懂这个由小区间-->大区间的结构，，还写不出来

还有那个dp初值问题，感觉dp的这个状态就和上一个状态有关系，赋值在这儿无所谓吧？

#include <iostream>#include <cstring>//#define inf 1<<20using namespace std;int dp[2500][2500],mincost[200];    char s[2500],c;int main(){    int n,m,add,del;    cin>>n>>m>>s;    for(int i=0;i<n;++i)    {        cin>>c>>add>>del;        mincost[c]=min(add,del);    }  //  memset(dp,inf,sizeof(dp));    for(int i=m-1;i>=0;--i)        for(int j=i+1;j<m;++j)    {        if(s[i]==s[j])            dp[i][j]=dp[i+1][j-1];        else        {            //dp[i][j]=min(dp[i+1][j]+mincost[s[i]],dp[i][j]);           // dp[i][j]=min(dp[i][j-1]+mincost[s[j]],dp[i][j]);    dp[i][j] = min(dp[i+1][j]+mincost[s[i]],dp[i][j-1]+mincost[s[j]]);        }    }    cout<<dp[0][m-1]<<endl;       return 0;}

代码2：

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define inf 1e8using namespace std;int dp[2020][2020];int main(){    int n,m,vis[200];    while(~scanf("%d%d",&n,&m))    {        char a[2020];        scanf("%s",a);        int x,y;        char z;        for(int i=0;i<n;i++)        {            cin>>z>>x>>y;            vis[z]=min(x,y);        }        for(int j=1;j<m;j++)        {            for(int i=j-1;i>=0;i--)                {                    if(a[i]==a[j])                        dp[i][j]=dp[i+1][j-1];                    else {                        dp[i][j]=min(dp[i+1][j]+vis[a[i]],dp[i][j-1]+vis[a[j]]);                    }                }        }        printf("%d\n",dp[0][m-1]);    }    return 0;}