POJ_3280_Cheapest_Palindrome_(动态规划)

描述

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http://poj.org/problem?id=3280

n 种小写字母构成长度为 m 的串,现在要通过增删字母使串回文,给出每种字母增和删的费用,求最小花费.

Cheapest Palindrome

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8040		Accepted: 3906

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

分析

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用 dp [ i ][ j ]表示使i~j区间回文的最小画费.

1.如果 a [ i ] == a [ j ] 那么要让 i ~ j 回文只需要处理 ( i+1 ) ~ ( j-1 )即可,所以 dp [ i ][ j ] = dp [ i+1 ][ j-1 ].

2.其他情况就用如下递推关系:

dp [ i ][ j ] = min ( dp [ i+1 ][ j ] + cost [ a [ i ] ] , dp [ i ][ j-1 ] + cost [ a[ j ] ] )

如果从回文的 ( i+1 ) ~ j 而来,添加了 a [ i ] 后就不回文了, 可以再在右边加一个 a[ i ] ,或者删掉左边的 a [ i ] ,所以 cost [ a [ i ] ] 取增删中最小的即可, i ~ ( j-1 ) 同理.

 #include<cstdio>
 #include<algorithm>
 #define read(a) a=getnum()
 using namespace std;

 const int maxn=,maxm=;
 int n,m;
 int a[maxm],cost[maxn];
 int dp[maxm][maxm];

 inline int getnum()
 {
     int ret=,k=;char c;
     for(c=getchar();c<''||c>'';c=getchar()) if(c=='-') k=-;
     for(;c>=''&&c<='';c=getchar()) ret=ret*+c-'';
     return ret*k;
 }

 void solve()
 {
     for(int i=m;i>=;i--)
     {
         for(int j=i;j<=m;j++)
         {
             if(a[i]==a[j]) dp[i][j]=dp[i+][j-];
             else dp[i][j]=min(dp[i+][j]+cost[a[i]],dp[i][j-]+cost[a[j]]);
         }
     }
     printf("%d\n",dp[][m]);
 }

 void init()
 {
     read(n); read(m);
     char c;
     for(int i=;i<=m;i++)
     {
         c=getchar();
         a[i]=c-'a';
     }
     getchar();
     for(int i=;i<=n;i++)
     {
         int c1,c2;
         c=getchar();
         read(c1); read(c2);
         cost[c-'a']=min(c1,c2);
     }
 }

 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("cheap.in","r",stdin);
     freopen("cheap.out","w",stdout);
 #endif
     init();
     solve();
 #ifndef ONLINE_JUDGE
     fclose(stdin);
     fclose(stdout);
     system("cheap.out");
 #endif
     return ;
 }