Cheapest Palindrome
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 7869 | Accepted: 3816 |
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.
题意：给定串，串中每个字符都有两种权值，代表删除它或者添加它所需要的花费。问将一个串变成回文串所需要的最小代价。
题解：区间DP，以前求解过这种题，不过是在回文串中添加字符求最少要添加多少字符变成回文串，那种题可以用求正反串的最长公共子序列然后用原串长度
减掉最长公共子序列长度得到。而这题多了个权值，明显就不能用那种方法了（看可以看测试用例）
这里的花费在删除和添加中选个小的就行了，因为删除和添加在本质上对回文串的作用相同。
这里的方法是区间DP：
回文串拥有很明显的子结构特征，即当字符串X是一个回文串时，在X两边各添加一个字符‘a‘后，aXa仍然是一个回文串，我们用d[i][j]来表示 A[i...j]这个子串变成回文串所需要添加的最少花费，那么对于A[i] == A[j]的情况，很明显有 d[i][j] = d[i+1][j-1] （这里需要明确一点，当i+1 > j-1时也是有意义的，它代表的是空串，空串也是一个回文串，所以这种情况下d[i+1][j-1] = 0）当A[i] != A[j]时，我们将它变成更小的子问题求解，我们有两种决策：
1、在A[j]后面添加一个字符A[i],花费为d[i+1][j]+value[i]；
2、在A[i]前面添加一个字符A[j],花费为d[i][j-1] + value[j]；
根据两种决策列出状态转移方程为：
d[i][j] = min{ d[i+1][j]+value[i], d[i][j-1]+value[j] } ; (每次状态转移，区间长度增加1)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include <string.h>
#include <math.h>
using namespace std;
int n,m;
char str[2000];
int value[26];
int dp[2000][2000]; ///代表从 i,j的最小花费
int main()
{
while(scanf("%d%d",&n,&m)!=EOF){
scanf("%s",str);
while(n--){
char c[2];
int v1,v2;
scanf("%s%d%d",c,&v1,&v2);
value[c[0]-‘a‘] = min(v1,v2); ///删掉和添加都可以变成回文串，选小的
}
int len = strlen(str);
memset(dp,0,sizeof(dp));
///起点i应当趋0，终点j应当趋len。
for(int i=len-1;i>=0;i--){
for(int j=i+1;j<len;j++){
if(str[i]!=str[j]) {
///更新区间[i ,j]选择[i+1,j]左边添加str[i]在[i,j-1]右边添加str[j]中小的那个
dp[i][j] = min(dp[i+1][j]+value[str[i]-‘a‘],dp[i][j-1]+value[str[j]-‘a‘]);
}else{
///相等的话直接添上去
dp[i][j]=dp[i+1][j-1];
}
}
}
printf("%d\n",dp[0][len-1]);
}
return 0;
}