hdu 5510 Bazinga(字符串kmp)

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.

For 

 

 

Input

The first line contains an integer 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;
}