hdu 5745（2016 Multi-University Training Contest 2）

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 528    Accepted Submission(s): 257

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

34 1abaca4 2aaaaaa9 3abcbacacbabc

 

Sample Output

10101110100100100

 

Author

zimpha

 

Source

2016 Multi-University Training Contest 2

题意：一个母串s，一个字串p，求p串是否能变成s的某一子串，p可以交换任意两个相邻字符，每个字符只能交换一次；最后得到一个二进制串，1表示s串从该位置为起始点，p变成与之一样；0反之。（据说比赛的时候好多人骂这题，说是条件1是可以挑两个字符交换......反正那句我没看懂，然后没管它....就那么AC了.....）

分析：我的想法是，先判断p和在s中对应的字串的某值（不知道怎么解释，因为是自己定义的.....）是否一样，不一样说明无论p怎么变都不可能变成s的那个字串，一样再继续，一位一位的判断，如果不一样就判断后一位是否和当前一样........总之也就是暴力一遍......

 

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 1e9;const int MOD = 1e9+7;#define ll long long#define CL(a,b) memset(a,b,sizeof(a))#define lson (i<<1)#define rson ((i<<1)|1)#define N 100010int n,m;int sum_s[N],sum_p[N];char s[N],p[5005];int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        scanf("%s%s",s+1,p+1);        sum_s[0] = sum_p[0] = 0;        for(int i=1; i<=n; i++)            sum_s[i] = sum_s[i-1]+s[i]-'a';        for(int i=1; i<=m; i++)            sum_p[i] = sum_p[i-1]+p[i]-'a';        bool flag;        int pp = sum_p[m];        for(int i=m; i<=n; i++)        {            int ss = sum_s[i]-sum_s[i-m];            flag = true;            if(ss == pp)            {                for(int j=i-m+1; j<=i; j++)                {                    if(s[j] != p[m+j-i])                    {                        if(j+1<=i&&s[j]==p[m+j-i+1]&&s[j+1]==p[m+j-i])                        {                            j++;                        }                        else {flag = false; break;}                    }                }            }            else flag = false;            if(flag) printf("1");            else printf("0");        }        for(int i=n-m+1; i<n; i++)            printf("0");        printf("\n");    }    return 0;}