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SDJZU_新生_字符串匹配（KMP）_A - Number Sequence
A - Number Sequence
Crawling in process...Crawling failedTime Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int next[10007];
int x[1000077];
int y[1000077];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int a,b,i;
scanf("%d%d",&a,&b);
for(i=0;i<a;i++)
{
scanf("%d",&x[i]);
}
for(i=0;i<b;i++)
{
scanf("%d",&y[i]);
}
next[0]=next[1]=0;
for(i=1;i<b;i++)
{
int j=next[i];
while(j&&y[i]!=y[j])
{
j=next[j];
}
next[i+1]=y[i]==y[j]?j+1:0;
}
int j=0,flag=0;
for(i=0;i<a;i++){
while(j&&x[i] != y[j])j = next[j];
if(x[i] == y[j])j++;
if(j == b){flag = 1;break;}
}
if(flag)printf("%d\n",i-b+2);
else printf("-1\n");
}
}
/*对于next那一块 ，直到现在感觉也不是很明白，这应该是核心了吧，只是照着我春哥的末班就写上了，现在仅仅是明白了KMP方法的思想了，却写不出来，只能通过做题慢慢磕了*/
/*后面这些代码是我春哥的,希望有助于以后的理解。*/
#include<stdio.h>
#include<string.h>
using namespace std;
int next[10010];
int aa[1000010];
void getNext(int *bb,int b){
next [0] = next[1] = 0;
for(int i=1;i<b;i++){
int j = next[i];
while(j&&bb[i]!=bb[j])j = next[j];
next[i+1] = bb[i] == bb[j]?j+1:0;
}
}
int main(){
int n,a,b,bb[10010];
scanf("%d",&n);
while(n--){
int flag = 0;
scanf("%d%d",&a,&b);
for(int i=0;i<a;i++){
scanf("%d",&aa[i]);
}
for(int i=0;i<b;i++){
scanf("%d",&bb[i]);
}
getNext(bb,b);
int j =0;
int i;
for(i=0;i<a;i++){
while(j&&aa[i] != bb[j])j = next[j];
if(aa[i] == bb[j])j++;
if(j == b){flag = 1;break;}
}
if(flag)printf("%d\n",i-b+2);
else printf("-1\n");
}
}SDJZU_新生_字符串匹配（KMP）_A - Number Sequence
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