题目链接:http://poj.org/problem?id=3280

Time Limit: 2000MS Memory Limit: 65536K

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:

给出n个字母,一个长度为m的字母串(字母都是从n个字母中挑);

然后给出每个字母的添加删除价格,表示添加一个或者删除一个该字母需要花费多少;

求把字母串变成一个回文串的最少花费;

题解:

设dp[0][m-1]为所求答案,且假设当我们求dp[i][j]时,所有的dp[ii][jj](i<ii<jj<j)都是已知的;

那么,有:

if(str[i]==str[j]) dp[i][j]=min(dp[i][j],dp[i+][j-]);
dp[i][j]=min(dp[i][j],dp[i][j-]+min(str[j].add,str[j].del));
dp[i][j]=min(dp[i][j],dp[i+][j]+min(str[i].add,str[i].del));

AC代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
struct Alpha{
char ch;
int add,del;
int mini(){return min(add,del);}
}alpha[];
int n,m;
char str[];
int dp[][]; int main()
{
cin>>n>>m;
cin>>str;
for(int i=;i<=n;i++)
{
char c; int add,del;
cin>>c>>add>>del;
alpha[c]=(Alpha){c,add,del};
//printf("%c %d %d\n",alpha[c].ch,alpha[c].add,alpha[c].del);
} memset(dp,INF,sizeof(dp));
for(int len=;len<=;len++)
{
for(int i=,j=i+len-;j<m;i++,j=i+len-) dp[i][j]=;
}
for(int len=;len<=m;len++)
{
for(int i=,j=i+len-;j<m;i++,j=i+len-)
{
if(str[i]==str[j]) dp[i][j]=min(dp[i][j],dp[i+][j-]);
dp[i][j]=min(dp[i][j],dp[i][j-]+alpha[str[j]].mini());
dp[i][j]=min(dp[i][j],dp[i+][j]+alpha[str[i]].mini());
}
} cout<<dp[][m-]<<endl;
}

POJ 3280 - Cheapest Palindrome - [区间DP]的更多相关文章

  1. POJ 3280 Cheapest Palindrome (区间DP) 经典

    <题目链接> 题目大意: 一个由小写字母组成的字符串,给出字符的种类,以及字符串的长度,再给出添加每个字符和删除每个字符的代价,问你要使这个字符串变成回文串的最小代价. 解题分析: 一道区 ...

  2. POJ 3280 Cheapest Palindrome ( 区间DP && 经典模型 )

    题意 : 给出一个由 n 中字母组成的长度为 m 的串,给出 n 种字母添加和删除花费的代价,求让给出的串变成回文串的代价. 分析 :  原始模型 ==> 题意和本题差不多,有添和删但是并无代价 ...

  3. POJ 3280 Cheapest Palindrome(DP 回文变形)

    题目链接:http://poj.org/problem?id=3280 题目大意:给定一个字符串,可以删除增加,每个操作都有代价,求出将字符串转换成回文串的最小代价 Sample Input 3 4 ...

  4. (中等) POJ 3280 Cheapest Palindrome,DP。

    Description Keeping track of all the cows can be a tricky task so Farmer John has installed a system ...

  5. POJ 3280 Cheapest Palindrome【DP】

    题意:对一个字符串进行插入删除等操作使其变成一个回文串,但是对于每个字符的操作消耗是不同的.求最小消耗. 思路: 我们定义dp [ i ] [ j ] 为区间 i 到 j 变成回文的最小代价.那么对于 ...

  6. POJ 3280 Cheapest Palindrome(DP)

    题目链接 题意 :给你一个字符串,让你删除或添加某些字母让这个字符串变成回文串,删除或添加某个字母要付出相应的代价,问你变成回文所需要的最小的代价是多少. 思路 :DP[i][j]代表的是 i 到 j ...

  7. POJ 3280 Cheapest Palindrome 简单DP

    观察题目我们可以知道,实际上对于一个字母,你在串中删除或者添加本质上一样的,因为既然你添加是为了让其对称,说明有一个孤立的字母没有配对的,也就可以删掉,也能满足对称. 故两种操作看成一种,只需要保留花 ...

  8. POJ 3280 Cheapest Palindrome (DP)

     Description Keeping track of all the cows can be a tricky task so Farmer John has installed a sys ...

  9. POJ 3280 Cheapest Palindrome(区间DP求改成回文串的最小花费)

    题目链接:http://poj.org/problem?id=3280 题目大意:给你一个字符串,你可以删除或者增加任意字符,对应有相应的花费,让你通过这些操作使得字符串变为回文串,求最小花费.解题思 ...

随机推荐

  1. 8 -- 深入使用Spring -- 2...3 使用@Resource配置依赖

    8.2.3 使用@Resource配置依赖 @Resource 位于javax.annotation包下,是来自Java EE规范的一个Annotation,Spring直接借鉴了该Annotatio ...

  2. javascript全屏操作

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  3. c 网络字节序和本机字节序转换

    将多字节整数类型的数据,从主机的字节顺序转化为网络字节顺序 #include <netinet/in.h> uint32_t htonl(uint32_t hostlong);uint16 ...

  4. dedeCMS解码

    var str = 'arrs1[]=99&arrs1[]=102&arrs1[]=103&arrs1[]=95&arrs1[]=100&arrs1[]=98& ...

  5. 管理工具 django-admin.py的相关命令列表

    C:\Users\lenovo> django-admin.py Type 'django-admin.py help <subcommand>' for help on a spe ...

  6. 执行RF设置顶层测试套件的描述说明

    场景1:通过pybot命令更新套件层描述 命令:pybot -D 套件层描述 -D --doc documentation 设置顶层测试套件的描述说明.说明中下划线将转换为空格, 并且他可能包含简单的 ...

  7. osgearth缓存数据命令

    新建一个.bat文件 中国地区 osgearth_package.exe ReadyMap.earth --tms ReadyMap.earth --out D:\APICenter\EarthDat ...

  8. Spring Web 应用的最大败笔

    开发人员在使用Spring应用是非常擅长谈论依赖注入的好处.不幸的是,他们不是那么真的利用它的好处,如单一职责原则,分离关注原则.如果我们一起来看看大部分Spring的Web应用程序,常见的错误的设计 ...

  9. ScaleType属性

    FIT_CENTER 把原图按照比例放大缩小到ImageView的高度,显示在ImageView的center(中部/居中显示). 1   2 CENTER_CROP 会拉伸图片以原图填满ImageV ...

  10. 《Lua程序设计》第5章 函数 学习笔记

    Lua为面向对象式的调用也提供了一种特殊的语法——冒号操作符.表达式o.foo(o, x)的另一种写法是o:foo(x),冒号操作符是调用o.foo时将o隐含地作为函数的第一个参数.Lua可以调用C语 ...