HDU5745 La Vie en rose

最新推荐文章于 2016-08-15 20:17:57 发布

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题目链接：HDU5745

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 871 Accepted Submission(s): 467

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

(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

Author

zimpha

Source

2016 Multi-University Training Contest 2

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wange2014 | We have carefully selected several similar problems for you: 5746 5743 5741 5740 5739

题意：比赛的时候看了2个小时没有理解题目意思。下来对着题解的代码凝视了好久终于大概明白了一些。题的意思是进行多模式串的匹配，在给定的模式串p上给每个位置标号1到lp，然后从1到p－1随意找出一组标号，要求标号要不能相连，就是隔着取，然后把这些标号对应的位置与相邻的右一个交换位置，这样就组成了一堆新的模式串。要求求出所有匹配的模式串的起始位置。

题目分析：按照答案的正解是dp加位操作，这个我过一阵再写，网上大多数做法是暴力搞，复杂度O（mn），4500ms勉强过。仔细想下10w*5k*T竟然不到5s就跑过了，数据也是够弱的。

//
//  main.cpp
//  La Vie en rose
//
//  Created by teddywang on 2016/7/21.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAX 500010
#define MIN 11000
int T,ls,lp;
int poss,posp;
int ans[100010];
char s[100010],p[5005];
int main()
{
    cin>>T;
    while(T--)
    {
        memset(ans,0,sizeof(ans));
        scanf("%d%d",&ls,&lp);
        scanf("%s",s);
        scanf("%s",p);
        poss=posp=0;
        while(poss<ls&&posp<lp)
        {
            if(s[poss]==p[posp])
            {
                if(posp==lp-1)
                {
                    ans[poss-posp]=1;
                    poss=poss-posp+1;
                    posp=0;
                }
                else
                {
                    poss++;
                    posp++;
                }
            }
            else if(s[poss]==p[posp+1]&&s[poss+1]==p[posp])
            {
                if(posp==lp-2)
                {
                    ans[poss-posp]=1;
                    poss=poss-posp+1;
                    posp=0;
                }
                else
                {
                    poss+=2;
                    posp+=2;
                }
            }
            else
            {
                poss=poss-posp+1;
                posp=0;
            }
        }
        for(int i=0;i<ls;i++)
            printf("%d",ans[i]);
        printf("\n");
    }
}