http://poj.org/problem?id=3280
http://www.hankcs.com/program/cpp/poj-3280-cheapest-palindrome.html
2.3 记录结果再利用的“动态规划” 基础的动态规划算法
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。
Read full article from 3280 -- 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.
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.
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
http://www.hankcs.com/program/cpp/poj-3280-cheapest-palindrome.html
2.3 记录结果再利用的“动态规划” 基础的动态规划算法
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。
定义dp[i][j]表示将原字串s的子字串s[i...j]变换成回文的最小花费，满足如下递推关系式：
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| 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]); } |
注意此处的可以增加也可以删除是个局，取其花费最小值即可。因为如果在首尾增删字母x都可以使一个字串s[i...j]变成回文的话，当然选取花费小的。
另一个注意点是两重循环的方向，起点i应当趋0，终点j应当趋M。
http://gisyhy.blog.163.com/blog/static/129390343200992485243739/
其实dp很难逃出3种思路：
1、一维线性dp：每次考虑i时，选择最优子问题要么在i-1，要么在1...i-1里；
2、二维线性dp：考虑(i,j)子问题时，选择最优子问题要么在(i+1,j)、(i,j-1)，要么在i<= k <=j，在k里；
3、树形dp：考虑i节点最优时，选择子节点最优，一般融合了01背包dp的双重dp。
上面3中模式也是我在做题后才发现的。
这个dp题其实就可以仿照第2中思路。
假设一个字符串Xx....yY；对于求这个字符串怎么求呢？
分4中情况讨论：
1、去掉X，取x....yY回文；
2、去掉Y，取Xx....y回文；
3、在左边加上X,取Xx....yYX回文；
4、在右边加上Y,取YXx....y回文。
至于去掉X、Y肯定没有第1、2中情况合算；加上X、Y肯定没有第3、4中情况合算。
因此令dp[i][j]为i...j要变成回文字符串的最小代价。
方程：
dp[i][j] = min{ dp[i+1][j] + {去掉X的代价}，dp[i+1][j] + {加上X的代价}，
dp[i][j-1]+ {去掉Y的代价}，dp[i][j-1] +{加上Y的代价}}；
其实分析发现，对于X而言，只要去 去掉 和加上 X 最小代价就行（因为前面dp串一样）,Y同理。
因此最后得出：
dp[i][j] = min{ dp[i+1][j] +min{ {去掉X的代价}, {加上X的代价}}，
dp[i][j-1]+min{ {去掉Y的代价}, {加上Y的代价}}}；
dp时候还有些注意事项：
比如当X和Y字符一样时，则在dp时必须先为x...y的最小代价。
int cost[32];int dp[2048][2048];///////////////////////////SubMain//////////////////////////////////int main(int argc, char *argv[]){#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout);#endif 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;#ifndef ONLINE_JUDGE fclose(stdin); fclose(stdout); system("out.txt");#endif return 0;}