poj 3280 Cheapest Palindrome

题目大意：一个字符串，用已知的字符把它变为回文串时，代价最小是多少？其中添加一个字符或删除一个字符都有相应代价。

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.

 

动态规划。

dp[i][j]表示由位置i到位置 j的子串变成回文串所需要的代价

如果str[i] == str[j] ，那么dp[i][j] = dp[i+1][j-1]

如果str[i] != str[j] , 那么 dp[i][j] = min(dp[i][j-1] + cost[str[j] - 'a'], dp[i+1][j] + cost[str[i] - 'a']);

即把i到j-1变为回文串的代价再加上加一个或减一个边缘字母的代价  和

  把i+1到j变为回文串的代价再加上加一个或减一个边缘字母的代价 的最小值

 

代码如下

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;

char str[2002];
int dp[2002][2002];
int cost[28];

int main() {
    int m, n;
    //freopen("input.txt","r",stdin);
    while(scanf("%d %d",&m,&n) != EOF) {
        scanf("%s",str);
        memset(cost, 0, sizeof(cost));
        for(int i = 0; i < m; i++) {
            int a, b;
            char c;
            cin >> c >> a >> b;
            cost[c-'a'] = min(a,b);
        }
        memset(dp, 0, sizeof(dp));
        for(int len = 1; len < n; len++) {
            for(int i = 0, j = len; i < n && j < n; i++,j++) {
                if(str[i] == str[j]) {
                    dp[i][j] = dp[i+1][j-1];
                }
                else {
                    dp[i][j] = min(dp[i][j-1] + cost[str[j] - 'a'], dp[i+1][j] + cost[str[i] - 'a']);
                }
            }
        }
        printf("%d
",dp[0][n-1]);
        
    }
    return 0;
}