2016多校联赛2D La Vie en rose（hdu 5745）

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 744    Accepted Submission(s): 386

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
  

 

Author

zimpha

 

Source

2016 Multi-University Training Contest 2

 

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wange2014   |   We have carefully selected several similar problems for you:   5746  5744  5743  5742  5741 

题意：p对s每一段子字符串进行匹配，如果p可以通过相邻两个交换得到s的子串，那么该s子串的首位就改为1，否则为0，注意相邻两个位置只能交换一次，例如第一位和第二位交换了，那么第二位不可以和第三位交换。其实题目意思我也不太懂，但是最后化简问题就是这样。下面给代码。

#include<iostream>
#include<stack>
#include<cstring>
#include<map>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<utility>
using namespace std;
#define maxn 100005
char s[maxn], p[5005];
int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		int slength, plength;
		scanf("%d%d", &slength, &plength);
		scanf("%s%s", s, p);
		for (int i = 0; i < slength; i++) {//从第一位开始匹配
			char x = '1';//x为第i个位置最后输出的数字
			for (int j = 0; j < plength;) {
				if (s[i+j] == p[j]) {
					j++;//如果这一位一样的话看下一位
				}
				else if (s[i + j] == p[j + 1] && s[i + j + 1] == p[j]) {
					j += 2;//如果相邻两位一样的话就跳到第三位
				}
				else {
					x = '0';//如果以上两种情况都不是，那么他不能匹配
					break;
				}
			}
			s[i] = x;
		}
		printf("%s\n", s);
	}
}