Number Sequence（kmp）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19246    Accepted Submission(s): 8267

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

题解：此题就是如果匹配就输出开始匹配时的数组下标；next数组有两个含义：位置还有长度；

让求串2在串1中首次出现的位置；

代码：

 #include<stdio.h>
 const int MAXN=;
 int a[MAXN*],b[MAXN],len1,len2,next[MAXN];
 void getnext(){
     int i=,j=-;
     next[i]=j;
     while(i<len2){
         if(j==-||b[i]==b[j]){
             i++;j++;
             next[i]=j;
         }
         else j=next[j];
     }
 }
 int kmp(){
     getnext();
     int i=,j=;
     while(i<len1){
         if(j==-||a[i]==b[j]){
             i++;j++;
             if(j==len2)return i-j+;
         }
         else j=next[j];
     }
     return -;
 }
 int main(){
     int T;
     int N,M;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&N,&M);
         for(int i=;i<N;i++)scanf("%d",&a[i]);
         for(int i=;i<M;i++)scanf("%d",&b[i]);
         len1=N;len2=M;
         printf("%d\n",kmp());
     }
     return ;
 }

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace  std;
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
int p[MAXN];
int N,M;
int s[MAXN];
int m[MAXN];
void getp(){
	int i=0,j=-1;
	p[0]=-1;
	while(i<M){
		if(j==-1||s[i]==s[j]){
			i++;j++;
			p[i]=j;
		}
		else j=p[j];
	}
}

void kmp(int& ans){
	getp();
	int j=0,i=0;
	while(i<N){
		if(j==-1||s[j]==m[i]){
			i++;j++;
			if(j==M){
				ans=i-j+1;
				return ;
			}
		}
		else j=p[j];
	}
}

int main(){
	int T;
	SI(T);
	while(T--){
		SI(N);SI(M);
		for(int i=0;i<N;i++)SI(m[i]);
		for(int i=0;i<M;i++)SI(s[i]);
		int ans=0;
		kmp(ans);
		if(!ans)puts("-1");
		else printf("%d\n",ans);
	}
	return 0;
}

　　str函数超时：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace  std;
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
char s1[MAXN],s2[MAXN];
int N,M;
/*
int s[MAXN];
int m[MAXN];
void getp(){
	int i=0,j=-1;
	p[0]=-1;
	while(i<M){
		if(j==-1||s[i]==s[j]){
			i++;j++;
			p[i]=j;
		}
		else j=p[j];
	}
}

void kmp(int& ans){
	getp();
	int j=0,i=0;
	while(i<N){
		if(j==-1||s[j]==m[i]){
			i++;j++;
			if(j==M){
				ans=i-j+1;
				return ;
			}
		}
		else j=p[j];
	}
}
*/
int main(){
	int T;
	SI(T);
	while(T--){
		SI(N);SI(M);
		int temp;
		for(int i=0;i<N;i++)SI(temp),s1[i]=temp+'0';
		for(int i=0;i<M;i++)SI(temp),s2[i]=temp+'0';
		s1[N]='\0';s2[M]='\0';
		int ans=0;
		ans=strstr(s1,s2)-s1;
		if(ans<0)puts("-1");
		else{
			printf("%d\n",ans+1);
		}
	}
	return 0;
}