POJ - 3280 Cheapest Palindrome 【区间dp】【非原创】

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.

 

原创博客：戳这里

题目大意：给定一个字符串S及其长度M与S所含有的字符种数N（最多26种小写字母），然后给定这N种字母Add与Delete的代价，求将S变为回文串的最小代价和。

 

分析：

这题算得上是一个很经典的模型了。先来说说最原始的模型，通过增添或者删除某些字母将一个字符串变为回文串，求最少增添或者删除次数，其实在这里增添和删除是没有区别的，我们只考虑删除就行了，对于这个模型通常有2种解法：

①将S反转得到S'，然后求两者最长公共子串的长度L，再用S的长度减去L便是答案了

②用 dp[i][j] 表示 [i,j] 区间的串的最优解，

若s[i]==s[j]，则dp[i][j]=dp[i+1][j-1]（请思考为什么）

否则 dp[i][j] = min (d[i+1][j] ， dp[i][j-1]) + 1，因为s[i]!=s[j]，所以必定要删去左右边界的某一个才能变成一个回文串，至于到底删去哪一个，当然是取 dp[i+1][j] 和 dp[i][j-1] 的MIN了

然后反观这个题，只是在add和delete的操作上附加了一个代价，所以这里的add和delete便不再等价了（当然是在add!=delete时），于是对于 [i+1,j] 和 [i,j] 之间的转换，当然要选择是在 [i+1,j] 上add一个字母变为 [i,j] 还是 在[i,j] 上delete一个字母变为 [i+1,j] 中的最优解了。

所以第一种方法显然不再适用，对于第二种方法，只要稍微改一下就行了

若s[i]==s[j]，则dp[i][j]=dp[i+1][j-1]仍然成立

否则 dp[i][j] = min  (   d[i+1][j] + min(add[s[i]],delete[s[i]])  ，   dp[i][j-1]  + min(add[s[j]],delete[s[j]]) )   注意写代码时别用min嵌套，会出错！

 

#include <iostream>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 2 * 1e3 + 10;
const int maxx = 1e6 + 10;
int dp[maxn][maxn];
string sa;
map<char, int> w;
int main()
{
    int n, m;
    cin >> n >> m;
    cin >> sa;
    char u;
    int x, y;
    for(int i = 0; i < n; ++i)
    {
        cin >> u >> x >> y;
        w[u] = min(x, y);
    }
 //   cout<<w[sa[0]]<<endl;
    for(int i = 1; i < m; ++i) //区间长度为i+1
    {
        for(int j = 0; i + j < m; ++j)
        {
            if(sa[j] == sa[i + j])
                dp[j][i + j] = dp[j + 1][i + j - 1];
            else
            {
                dp[j][i + j] = min(dp[j + 1][i + j] + w[sa[j]], dp[j][i + j - 1] + w[sa[i + j]]);

            }
        }
    }
    cout<< dp[0][m - 1] << endl;
    return 0;
}