題目鏈接：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?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? Recommend wange2014???|???We have carefully selected several similar problems for you:??5746?5743?5741?5740?5739?

題意：比賽的時候看了2個小時沒有理解題目意思。下來對著題解的代碼凝視了好久終于大概明白了一些。題的意思是進行多模式串的匹配，在給定的模式串p上給每個位置標號1到lp，然后從1到p－1隨意找出一組標號，要求標號要不能相連，就是隔著取，然后把這些標號對應的位置與相鄰的右一個交換位置，這樣就組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。

題目分析：按照答案的正解是dp加位操作，這個我過一陣再寫，網上大多數做法是暴力搞，復雜度O（mn），4500ms勉強過。仔細想下10w*5k*T竟然不到5s就跑過了，數據也是夠弱的。 // // 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");} }

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