省錢構建迴文串

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

題意:
給出一個字串s,通過增加或者刪除字母使得字串s成為一個迴文串
輸入中會給出每個字母增加或者減少的費用
然後求出最少費用使得字串成為一個迴文串
注意:空串也是一個迴文串

思路:
先理解選擇問題：
{
1.對於每個字母,根據費用選擇添還是刪
比如:abcb
第一步:a!=b,兩種選擇abcba和bcb,因此只需考慮是添a還是刪a
根據輸入自然選擇添a,dp[2][3]=1000;

確定好字母是添還是刪,假定我們選a為添,b也為添 

2.下面面臨下一種選擇
比如:abcb
a!=b,兩種選擇abcba和babcb,這兩種都可是第一個和最後一個相等的,取最小

}

dp[i][j]表示字串a從i到j構成迴文串的最小費用

dp[i][j]=min(dp[i+1][j]+s[(int)a[i]],dp[i][j-1]+s[(int)a[j]]);
利用該狀態轉移方程可求字串a從i到j構成迴文串的最小費用、
具體過程不是很明白,多模擬模擬幾遍
比如吧:求dp[1][3]=min(dp[2][3]+s[1],dp[1][2]+s[3]);
唉,就理解這麼多吧

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[2020];
int dp[2020][2020];//dp[i][j]表示字串a從i到j構成迴文串的最小費用 
int s[128];
int main()
{
	int m,n;
	while(~scanf("%d %d",&m,&n))
	{
		int i,j,s1,s2;
		char x;
		memset(dp,0,sizeof(dp));
		scanf("%s",a);
		for(i=0;i<m;i++)
		{
			getchar();
			scanf("%c %d %d",&x,&s1,&s2);
			s[(int)x]=min(s1,s2);
		}
		for(i=n-2;i>=0;i--)
		{
			for(j=i+1;j<n;j++)
			{//例如bc,我們要考慮的是變為cbc還是bcb         
				dp[i][j]=min(dp[i+1][j]+s[(int)a[i]],dp[i][j-1]+s[(int)a[j]]);
				if(a[i]==a[j])//當相等情況發生時,說明上面+s[]便是多加的費用,可能需要改變dp[i][j]的值 
					dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
			}
		}
		printf("%d\n",dp[0][n-1]);
	}
}