Number Sequence（kmp）
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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：
1 #include<stdio.h> 2 const int MAXN=10010; 3 int a[MAXN*100],b[MAXN],len1,len2,next[MAXN]; 4 void getnext(){ 5 int i=0,j=-1; 6 next[i]=j; 7 while(i<len2){ 8 if(j==-1||b[i]==b[j]){ 9 i++;j++; 10 next[i]=j; 11 } 12 else j=next[j]; 13 } 14 } 15 int kmp(){ 16 getnext(); 17 int i=0,j=0; 18 while(i<len1){ 19 if(j==-1||a[i]==b[j]){ 20 i++;j++; 21 if(j==len2)return i-j+1; 22 } 23 else j=next[j]; 24 } 25 return -1; 26 } 27 int main(){ 28 int T; 29 int N,M; 30 scanf("%d",&T); 31 while(T--){ 32 scanf("%d%d",&N,&M); 33 for(int i=0;i<N;i++)scanf("%d",&a[i]); 34 for(int i=0;i<M;i++)scanf("%d",&b[i]); 35 len1=N;len2=M; 36 printf("%d\n",kmp()); 37 } 38 return 0; 39 }
#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;
}
