（POJ 3280）Cheapest Palindrome <DP> 回文数变形

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 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.
Source

USACO 2007 Open Gold

题意：
字串S长M，由N个小写字母构成。欲通过增删字母将其变为回文串，增删特定字母花费不同，求最小花费。

分析：
对于这一题的分析，推荐先看看它的简化版：http://blog.csdn.net/stillxjy/article/details/54090847

看完简化版后，我们来说说这题
这题的不同之处在于它有了删除的操作。
仔细想想，其实删除的操作和添加的操作的等价的，因为你删除某个字符，其实等价于你在它的对称位置添加了一个数。
所以我们可以像简化版的题目一样只考虑添加，而操作某个字符s的费用为 min(添加s，删除s)

所以我们的公式为：

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

要注意循环的顺序： 在计算dp[i][j] 时我们要 考虑到dp[i][j-1],dp[i-1][j] ,dp[i+1][j-1] 所以决定了j 的顺序是正序的而i 的顺序的倒序的

AC代码：

#include <iostream>#include <string>using namespace std;int cost[32];int dp[2048][2048];int main(int argc, char *argv[]){    int N, M;    cin >> N >> M;    string s;    cin >> s;    for (int i = 0; i < N; ++i)    {        char c;        int add_cost, delete_cost;        cin >> c >> add_cost >> delete_cost;        cost[c - 'a'] = min(add_cost, delete_cost);    }    for (int i = M - 1; i >= 0; --i)    {        for (int j = i + 1; j < M; ++j)        {            dp[i][j] = min( dp[i + 1][j] + cost[s[i] - 'a'],    // 比i, j少了一个首字母                            dp[i][j - 1] + cost[s[j] - 'a']);   // 比i, j少了一个尾字母            if (s[i] == s[j])            {                // 首尾相同，等于少了首尾                dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);            }        }    }    cout << dp[0][M - 1] << endl;    return 0;}