【POJ3461】Oulipo

清疚 2022-09-25 07:22 145阅读 0赞

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      Oulipo 
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        <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
        <td>&nbsp;</td> 
        <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
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        <td><strong>Total Submissions:</strong>&nbsp;35904</td> 
        <td>&nbsp;</td> 
        <td><strong>Accepted:</strong>&nbsp;14492</td> 
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    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Description</p> 
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     <p>The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter&nbsp;'e'. He was a member of the Oulipo group. A quote from the book:</p> 
     <blockquote> 
      <p>Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…</p> 
     </blockquote> 
     <p>Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive&nbsp;'T's is not unusual. And they never use spaces.</p> 
     <p>So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet { 'A',&nbsp;'B',&nbsp;'C', …,&nbsp;'Z'} and two finite strings over that alphabet, a word&nbsp;<em>W</em>&nbsp;and a text&nbsp;<em>T</em>, count the number of occurrences of&nbsp;<em>W</em>&nbsp;in&nbsp;<em>T</em>. All the consecutive characters of W must exactly match consecutive characters of&nbsp;<em>T</em>. Occurrences may overlap.</p> 
    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Input</p> 
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     <p>The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:</p> 
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      <li>One line with the word&nbsp;<em>W</em>, a string over { 'A',&nbsp;'B',&nbsp;'C', …,&nbsp;'Z'}, with 1 ≤ |<em>W</em>| ≤ 10,000 (here |<em>W</em>| denotes the length of the string&nbsp;<em>W</em>).</li> 
      <li>One line with the text&nbsp;<em>T</em>, a string over { 'A',&nbsp;'B',&nbsp;'C', …,&nbsp;'Z'}, with |<em>W</em>| ≤ |<em>T</em>| ≤ 1,000,000.</li> 
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    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Output</p> 
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     <p>For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word&nbsp;<em>W</em>&nbsp;in the text&nbsp;<em>T</em>.</p> 
    </div> <p style="font-size:18pt; font-weight:bold; color:blue">Sample Input</p> <pre style="font-family:'Courier New',Courier,monospace; font-size:12pt">3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN</pre> <p style="font-size:18pt; font-weight:bold; color:blue">Sample Output</p> <pre style="font-family:'Courier New',Courier,monospace; font-size:12pt">1
3
0</pre> <p style="font-size:18pt; font-weight:bold; color:blue">Source</p> 
    <div style="font-family:'Times New Roman',Times,serif; font-size:12pt"> 
     <a href="http://poj.org/searchproblem?field=source&amp;key=BAPC+2006+Qualification" style="text-decoration:none" rel="nofollow">BAPC 2006 Qualification</a> 
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\[[Submit][]\] \[Go Back\] \[[Status][]\] \[[Discuss][]\]

这个是子串匹配，和模板的模式串匹配有一点不同，就那一点next数组值的存储范围包括第lens个，因为你判断玩一个子串后要判断是不是前面有已经重合的部分，就像第二个输入输出样例。

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int N = 1e6+20;
    const int M = 1e5+10;
    char p[N],s[M];
    int next[M];
    int lens,lenp;
    void getNext() {
    	int j,k;
    	next[0]=-1;
    	j=0,k=-1;
    	while(j!=lens) {
    		if(k==-1||s[k]==s[j]) {
    			k++;
    			j++;
    			next[j]=k;
    		} else
    			k=next[k];
    	}
    }
    int KMPMatch() {
    	int i=0,j=0,count=0;
    	while(i<lenp) {
    		if(j==-1||p[i]==s[j]) {
    			i++;
    			j++;
    		} else
    			j=next[j];
    		if(j==lens) {
    			count++;
    			j=next[j];
    		}
    	}
    	return count;
    }
    int main() {
    	int T;
    	scanf("%d",&T);
    	while(T--) {
    		scanf("%s%s",s,p);
    		lens=strlen(s);			//加了这两句就能ac，不加直接strlen（s）就超时，我也是醉了 
    		lenp=strlen(p);			//不造为啥，以后要规范细节啊，否则容易吃大亏，
    								//尽量避免意向不到的错误而导致改都不知道怎么改 
    		getNext();
    		int ans=KMPMatch();
    		printf("%d\n",ans);
    	}
    	return 0;
    }

[题目地址：http://poj.org/problem?id=3461][http_poj.org_problem_id_3461]

[Submit]: http://poj.org/submit?problem_id=3461
[Status]: http://poj.org/problemstatus?problem_id=3461
[Discuss]: http://poj.org/bbs?problem_id=3461
[http_poj.org_problem_id_3461]: http://poj.org/problem?id=3461