标签：The 第一个 put main bsp 一个开始 ++ with style
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.
3 4 abcb a 1000 1100 b 350 700 c 200 800
900
#include <stdio.h> #include <iostream> #include <algorithm> #include <string.h> using namespace std; char a[2005]; int hashs[30];//字符的最小花费进行映射 int dp[2005][2005];//从某一字符到另一字符的组成的子字符串变成回文字符串所需要的花费 int main(void) { int add, del; char c; int n, m; scanf("%d %d", &n, &m); scanf("%s", a); for(int i = 0; i < n; i++) { while(scanf("%c", &c) && !(c <= ‘z‘ && c >= ‘a‘)) {} scanf("%d %d", &add, &del);//读入时计算到底是删还是加划算 hashs[c - ‘a‘] = min(add, del); } for(int i = 1; i < m; i++)//i意味着起始位置到结束位置的差值 for(int j = 0; j + i < m; j++) //j意味着字符串的起始位置 { if(a[j] == a[j + i])//说明我们并不需要j + 1~j + i或者是j ~ j + i - 1组成的字符串变成回文所需要的花费，因为这样多算了本来不需要的一个字符的操作 dp[j][j + i] = dp[j + 1][j + i - 1];//如果i==j或者是i + 1 == j那么j + 1 > j + i - 1那么那个值理论上不该有，但是我们默认初始化为0，所以用上了。 else dp[j][j + i] = min(dp[j + 1][j + i] + hashs[a[j] - ‘a‘], dp[j][j + i - 1] + hashs[a[j + i] - ‘a‘]);//取对新字符串第一个进行操作划算还是对最后一个 } printf("%d\n", dp[0][m - 1]);//从原字符串的第一个开始到最后一个，使得该子串（其实也就是原字符串）变为回文字符串的最小花费 return 0; }
挑战程序设计竞赛2.3习题：Cheapest Palindrome POJ - 3280
标签：The 第一个 put main bsp 一个开始 ++ with style
原文地址：https://www.cnblogs.com/jacobfun/p/12229149.html
