poj3280Cheapest Palindrome

Cheapest Palindrome

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8864 Accepted: 4299

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 4abcba 1000 1100b 350 700c 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.

1.将字符串变为回文串分为子问题，就成为先将(i,j)区间的字符串变为回文，然后依次向外区间扩展,显然区间(i,j)与周边区间的关系可以参考LCS；

用dp[i][j]表示将区间(i,j)字符变为回文串所需的最小代价，则有递推公式

            if (s[i] == s[j]) {                dp[i][j] = dp[i+1][j-1];//与(i+1,j-1)            }else {                dp[i][j] = min(dp[i][j],dp[i+1][j]+min(add[i],del[i]));//与(i+1,j)关系，对s[i]处理                dp[i][j] = min(dp[i][j],dp[i][j-1]+min(add[j],del[j]));//与(i,j-1)关系，对s[j]处理            }

2.找到递推公式后，开始进行dp，两个for循环该怎么考虑呢

注意到i是从i+1状态到i，因此i应该是从大到小循环；

j是从j-1到j状态，因此j应该是从达到小循环，这样才能保证前面状态的结果成功应用到后面状态的递推中；

可以从字符串前面开始dp，这样的话j为外循环不断++，i不断--;

也可以从字符串后面开始，这样的话i为外循环不断--,j不断++；
我采用的是第二种

/** Filename:    code.cpp* Created:     2016-10-18* Author:        wyl6 *[mail:[e-mail]]* Desciption:  Desciption*/#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INT_MAX 1 << 30#define MAX 100#define mem(a) memset(a,0,sizeof(a))typedef long long ll;int n,m;int add,del;//花销char cha;//字母char s[2002];int cost[100];int dp[2002][2002];//表示从i到j改为回文需要的最小代价void solve(){    mem(dp);    for (int i = m-2; i >= 0; i -= 1){        for (int j = i+1; j < m; j += 1){            if (s[i] == s[j]) dp[i][j] = dp[i+1][j-1];            else dp[i][j] = min((dp[i+1][j]+cost[s[i]-'a']),(dp[i][j-1]+cost[s[j]-'a']));        }    }    printf("%d\n",dp[0][m-1]);}int main(int argc, char const* argv[]){    scanf("%d%d",&n,&m);    scanf("%s%*c",&s);//删去回车符    for (int i = 0; i < n; i += 1){        scanf("%c%d%d%*c",&cha,&add,&del);//删去回车符        cost[cha - 'a'] = min(add,del);    }    solve();    return 0;}