HDU 5745 La Vie en rose（DP，枚舉）

Problem Description

Professor Zhang would like to solve the multiple pattern matching problem, but he only has only one pattern string p=p1p2...pm. So, he wants to generate as many as possible pattern strings from p using the following method:

1. select some indices i1,i2,...,ik such that 1≤i1<i2<...<ik<|p| and |ij−ij+1|>1 for all 1≤j<k.
2. swap pij and pij+1 for all 1≤j≤k.

Now, for a given a string s=s1s2...sn, Professor Zhang wants to find all occurrences of all the generated patterns in s.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n≤105,1≤m≤min{5000,n}) -- the length of s and p.

The second line contains the string s and the third line contains the string p. Both the strings consist of only lowercase English letters.

 

Output

For each test case, output a binary string of length n. The i-th character is "1" if and only if the substring sisi+1...si+m−1 is one of the generated patterns.

 

Sample Input

3 4 1 abac a 4 2 aaaa aa 9 3 abcbacacb abc

 

Sample Output

1010 1110 100100100

 

題意：給定一個長度爲n的匹配串s和長度爲m的模式串p，p可以進行變換，變換規則是任意交換兩個相鄰的字符，但是每個字符最多（被）交換一次。結果輸出一個n位的二進制，對於匹配串的字符位置i，如果s的子串（si,si+1,si+2,...,si+m-1）可以由p串變換得出的話，則這個位置輸出“1”，否則輸出“0”

因爲每個字符最多交換或者被交換一次，那麼所有交換都是互不影響的，判斷p是不是可以構造成目標串，只需O(m)地遍歷一遍p串和對應位置的s串，遇到不可交換且不匹配就退出並輸出0，否則全部匹配就輸出1，然後O(n)地枚舉s的所有子串，即可在O(nm)時間得出答案（顯然i>n - m就直接輸出0），嚴格上說這不算DP，但題解上說這是O(nm)的DP，所以也就是運用了一些dp的簡單思想而已，代碼如下：

#include <bits/stdc++.h>

using namespace std;

char s[100010];
char p[10010];

int main()
{
	int t,n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&m);
		scanf("%s %s",s,p);
		for(int i = 0 ; i < n ; i++)
		{
			if(i > n - m)
			{
				putchar('0');
				continue;
			}
			bool f = true;
			for(int j = 0 ; j < m ; j++)
			{
				if(j < m - 1 && p[j] == s[i + j + 1] && p[j + 1] == s[i + j])
					j++;
				else if(p[j] != s[i + j])
				{
					f = false;
					break;
				}
			}
			putchar(f ? '1' : '0');
		}
		puts("");
	}

    return 0;
}