hdu 5510 Bazinga(字符串kmp)

时间：2023-03-08 21:37:29

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2287    Accepted Submission(s): 713

Problem Description
Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.
hdu 5510 Bazinga(字符串kmp)
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.

A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".

Input
The first line contains an integer t (1≤t≤50) which is the number of test cases.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
Output
For each test case, output the largest label you get. If it does not exist, output −1.
Sample Input
4
5
ab
abc
zabc
abcd
zabcd
4
you
lovinyou
aboutlovinyou
allaboutlovinyou
5
de
def
abcd
abcde
abcdef
3
a
ba
ccc
Sample Output
Case #1: 4
Case #2: -1
Case #3: 4
Case #4: 3
/*

hdu 5510 Bazinga(字符串kmp)

problem:

给你n个字符串,求一个最大的序号i. 使1~i-1的字符串中有一个不是str[i]的子串

solve:

最开始看见想的是字典树.但是感觉很麻烦,目测会超时.

然后想的是kmp,但是优化有问题.  每次倒着往前搜,想的是如果1~i-1的所有都是i的子串. 那么搜索到i时就可以直接返回true.

如果不匹配就退出,否则就一直搜索下去. 结果TLE了....很悲伤

后来发现正确优化:如果i是k的子串. 那么匹配的时候就不需要匹配i,因为如果当前与k匹配,那么就一定与i匹配....

所以过程中标记一下不用匹配的就行了

hhh-2016-08-29 21:00:19

*/

#pragma comment(linker,"/STACK:124000000,124000000")

#include <algorithm>

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <vector>

#include <math.h>

#include <queue>

#include <set>

#include <map>

#define lson  i<<1

#define rson  i<<1|1

#define ll long long

#define clr(a,b) memset(a,b,sizeof(a))

#define scanfi(a) scanf("%d",&a)

#define scanfs(a) scanf("%s",a)

#define scanfl(a) scanf("%I64d",&a)

#define key_val ch[ch[root][1]][0]

#define inf 1e9

using namespace std;

const ll mod = 1e9+7;

const int maxn = 2005;

char str[505][2004];

int flag[maxn];

void pre_kmp(char x[],int m,int kmpnext[])

{

    int i,j;

    j = kmpnext[0] = -1;

    i = 0;

    while(i < m)

    {

        while(j != -1 && x[i] != x[j])

            j = kmpnext[j];

        if(x[++i] == x[++j])

            kmpnext[i] = kmpnext[j];

        else

            kmpnext[i] = j;

    }

}

int nex[2005];

int kmp(char x[],char y[])

{

    int m = strlen(x);

    int n = strlen(y);

    int i,j;

    clr(nex,0);

    pre_kmp(x,m,nex);

    i = j = 0;

    while(i < n)

    {

        while(j != -1 && y[i] != x[j]) j = nex[j];

        i++,j++;

        if(j >= m)

        {

            return true;

        }

    }

    return false;

}

int main()

{

    int T,n;

//    freopen("in.txt","r",stdin);

    scanfi(T);

    int cas = 1;

    while(T--)

    {

        clr(flag,0);

        scanfi(n);

        int ans = -1;

        for(int i = 1; i <= n; i++)

        {

            scanfs(str[i]);

            for(int j =1;j < i;j++)

            {

                if(flag[j])

                    continue;

                if(kmp(str[j],str[i]))

                    flag[j] = 1;

                else

                {

                    ans = i;

                }

            }

        }

        printf("Case #%d: %d\n",cas++,ans);

    }

    return 0;

}

　　

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