POJ 3280 Cheapest Palindrome
来源:互联网 发布:2017淘宝骗保师 编辑:程序博客网 时间:2024/06/03 21:35
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 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
Sample Input
3 4abcb
a 1000 1100
b 350 700
c 200 800
Sample Output
900Hint
Source
回文串上的插入和删除字符其实是等价的,只要取价格最少的就可以了。。。。。
#include <cstdio>
#include <cstring>
using namespace std;
int n,m,dp[3000][3000],cost[30];
char str[3000];
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
scanf("%s",str);
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
char c; int p1,p2;
getchar();
scanf("%c%d%d",&c,&p1,&p2);
cost[c-'a']=min(p1,p2);
}
for(int i=1;i<m;i++)
{
for(int j=0;j+i<m;j++)
{
dp[j][i+j]=min(dp[j+1][i+j]+cost[str[j]-'a'],dp[j][i+j-1]+cost[str[i+j]-'a']);
if(str[j]==str[i+j])
dp[j][i+j]=min(dp[j][i+j],dp[j+1][i+j-1]);
}
}
printf("%d\n",dp[0][m-1]);
}
return 0;
}
- Poj 3280 Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- POJ 3280--Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- POJ - 3280 Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- ***POJ 3280 Cheapest Palindrome
- poj-3280-Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- POJ 3280 Cheapest Palindrome
- poj 3280 Cheapest Palindrome
- poj 3280Cheapest Palindrome
- POJ 3267 The Cow Lexicon
- Asp.Net MVC4入门指南(9):查询详细信息和删除记录
- DataGuard小结--1
- POJ 1054 The Troublesome Frog
- Asp.Net MVC4入门指南(10):第三方控件Studio for ASP.NET Wijmo MVC4 工具应用
- POJ 3280 Cheapest Palindrome
- POJ 1191 棋盘分割
- 2013 北京 QCon热点分享
- OpenCV学习笔记[1]初探OpenCV
- POJ 1260 Pearls
- Spread Studio中文支持图解
- 自定义取值距离的javascript random()函数
- HDOJ 3555 Bomb
- Spread for ASP.NET 7新功能使用指南