poj 3280 Cheapest Palindrome (区间dp)

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：

给出很一个含有m个字符的字符串，然给出字符串中出现的字符对应删除和添加的价值，问怎么操作使这个字符串成为回文，并且价值最小。

题解：

以前题解：​​点这里​​

思维好一点，省略了区间长度枚举，下面代码好想，而且也比较规范化。

状态：dp[i][j] 表示i到j的最小价值，这样思考：三种情况：

1、如果dp[i+1][j]已经是回文了，考虑是将i这个位的字符删除还是在j后面加上a[i] 

2、如果dp[i][j-1]是回文串了，其他的同上。

3、dp[i+1][j-1]是回文串了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
typedef long long lld;
#define INF 0x3f3f3f3f
#define maxn 2000+5
#define maxm 200
int dp[maxn][maxn];
char str[maxn];
int ist[maxm],del[maxm];
 
int min(int a,int b,int c)
{
    return min(a,min(b,c));
}
 
int main()
{
    int n,m,w1,w2;
    char c;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        scanf("%s",str+1);
        for(int i=1;i<=n;i++)
        {
             scanf("%*c%c %d %d",&c,&w1,&w2);
             ist[c]=w1;
             del[c]=w2;
        }
        memset(dp,0,sizeof dp);
        for(int L=2;L<=m;L++)
            for(int i=1;i+L-1<=m;i++)
            {
                int j=i+L-1;
                dp[i][j]=INF;
                dp[i][j]=min(dp[i][j],dp[i+1][j]+ist[str[i]],dp[i+1][j]+del[str[i]]);
                dp[i][j]=min(dp[i][j],dp[i][j-1]+ist[str[j]],dp[i][j-1]+del[str[j]]);
                if(str[i]==str[j])
                    dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
 
            }
        printf("%d\n",dp[1][m]);
    }
    return 0;
}