題意：?
給出一個字符串，要求將其修改成一個回文字符串，給出修改某種字母（添加或刪除）的價值，求最小使其成為回文字符串的價值。?
題解：?
感覺是求最長回文子序列的變形，然而剛開始想著用類似于求最長回文子序列的方法，將字符串反轉后與原字符串匹配最長公共子序列，在匹配的過程中來進行dp轉移，但發現這樣不行，因此還是借鑒了網上的方法用dp[i][j]表示處理完區間將ｉ～ｊ之間的字符串變為回文字符串的最小代價，那么很明顯有轉移方程

Time limit? ? ?2000 ms

Memory limit? ? ?65536 kB

OS? ? ? Linux

Source? ? ??USACO 2007 Open Gold

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.

題意：有n種字符，組成長為m的串，分別給出字符增加和刪除操作 的代價，求把原字符串變成回文串的最小代價

解題思路：區間dp.dp[i][j]表示從區間i到j是一個回文串的最小代價 ，狀態轉移方程是：如果s[i]=s[j]，dp[i][j]=dp[i+1][j-1]；否則 dp[i][j]=min(dp[i][j],dp[i+1][j]+min(add[s[i]-'a'],del[s[i]-'a'])))， dp[i][j]=min(dp[i][j],dp[i][j-1]+min(add[s[j]-'a'],del[s[j]-'a'])). 區間dp就是將一個區間劃分成很多個小區間進行求解

還有一點就是其實雖然給出的刪除和修改是兩個費用，但實際上每次修改時添加和刪除的效果是一樣的，因此每次都一定會取其中最小的費用，新開一個數組存起來就不用每次比較了。

#include <iostream> #include <algorithm> #include <cstring> #include <string> #include <cstdio> #include <cmath> using namespace std;#define INF 0x3f3f3f3f const int maxn=30; const int maxm=2000+5; int n,m; string s; int add[maxn],del[maxn]; int dp[maxm][maxm]; void solve() {int ans=0;for(int i=m-1;i>=0;i--){dp[i][i]=0;for(int j=i+1;j<m;j++){dp[i][j]=INF;if(s[i]==s[j])dp[i][j]=dp[i+1][j-1];else{dp[i][j]=min(dp[i][j],dp[i+1][j]+min(add[s[i]-'a'],del[s[i]-'a']));dp[i][j]=min(dp[i][j],dp[i][j-1]+min(add[s[j]-'a'],del[s[j]-'a']));}}}printf("%d\n",dp[0][m-1]); } int main() {while(cin>>n>>m){cin>>s;char ch;int a,b;for(int i=0;i<n;i++){cin>>ch>>a>>b;add[ch-'a']=a;del[ch-'a']=b;}solve();}return 0; }

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