动态规划之微软2017秋季校招B题

原题：

Alice writes an English composition with a length of N characters. However, her teacher requires that M illegal pairs of characters cannot be adjacent, and if 'ab' cannot be adjacent, 'ba' cannot be adjacent either.

In order to meet the requirements, Alice needs to delete some characters.

Please work out the minimum number of characters that need to be deleted.

输入

The first line contains the length of the composition N.

The second line contains N characters, which make up the composition. Each character belongs to 'a'..'z'.

The third line contains the number of illegal pairs M.

Each of the next M lines contains two characters ch1 and ch2,which cannot be adjacent.  

For 20% of the data: 1 ≤ N ≤ 10

For 50% of the data: 1 ≤ N ≤ 1000  

For 100% of the data: 1 ≤ N ≤ 100000, M ≤ 200.

输出

One line with an integer indicating the minimum number of characters that need to be deleted.

样例提示

Delete 'a' and 'd'.

这道题刚开始没想到是动态规划，后来才想到。

状态： f[i][j] 表示从0到i的子串，最后留下的字符串的最后一个字母是j，最后留下的最长字符串的长度

状态转移，逐个枚举进行了（思想可能不成熟，但总算是AC了，也记录一下动态规划的使用）

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Composition {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int stringLength, unAdjacent;
		String string;
		List pattern = new ArrayList();
		String temp;
		while(scanner.hasNext()){
			stringLength = scanner.nextInt();
			scanner.nextLine();
			string = scanner.nextLine();
			unAdjacent = scanner.nextInt();
			scanner.nextLine();
			pattern.clear();
			for (int i = 0; i < unAdjacent; i++) {
				temp = scanner.nextLine();
				pattern.add(temp);
				pattern.add(temp.charAt(1)+ "" + temp.charAt(0));
			}
			getMaxLength(string.toCharArray(), pattern);
		}
		scanner.close();
	}
	public static void getMaxLength(char[] string,  List pattern){
		//res[i][j]表示0...i的子串中留下的最后一个字母是j时的最大长度
		int[][] res = new int[string.length + 1][26];
		for(int j = 0; j < res[0].length; j++){
		}
		res[1][string[0] - 'a'] = 1; 
		for(int i = 2; i < res.length; i++){
			int maxVal = 1;
			for(int j = 0; j < res[0].length; j++){
				if(j != string[i - 1] - 'a'){
					res[i][j] = res[i - 1][j];
					continue;
				}
				for (int k = 0; k < res[0].length; k++) {
					if(res[i - 1][k] == 0){
						continue;
					}
					if (pattern.contains((char) (k + 'a') + "" + (string[i - 1]))) {
						continue;
					}
					if(res[i - 1][k] + 1 > maxVal){
						maxVal = res[i - 1][k] + 1;
					}
				}
				res[i][string[i - 1] - 'a'] = maxVal;
			}
		}
		int max = 0;
		for (int j = 0; j < res[0].length; j++) {
			max = Math.max(max, res[res.length - 1][j]);
		}
		System.out.println(string.length - max);
	}
}