hdu 1711

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20330 Accepted Submission(s): 8719

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

就是把KMP的字符换成了数组，注意ans不是-1 的时候要加1，

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN=1000005;int nxt[MAXN];int T[MAXN],W[MAXN]; //  T主串    W子串int ans,len1,len2;void get_nxt(int* s){    int i=0,j=-1;    nxt[0]=-1;    while(i<len2){        if(j==-1||s[i]==s[j]){            i++;            j++;            nxt[i]=j;        }        else {            j=nxt[j];        }    }}void kmp(int *s,int *t){//s主串，t子串    get_nxt(t);    int i=0,j=0;    int flag=0;    while(i<len1&&j<len2){        if(j==-1||s[i]==t[j]){            i++,j++;        }        else {            j=nxt[j];        }        if(len2==j){//   如果算可以重叠的子串的个数，ans++;j=nxt[j];//        ans=i-len2;  //不能重叠的个数，ans++;j=-1;//        flag=1;        break;    }    }    if(!flag){        ans=-1;    }}int main(){    int kase;    scanf("%d",&kase);    while(kase--){        scanf("%d%d",&len1,&len2);        for(int i=0;i<len1;i++)scanf("%d",T+i);        for(int i=0;i<len2;i++)scanf("%d",W+i);        kmp(T,W);        if(ans==-1)printf("-1\n");       else printf("%d\n",ans+1);    } return 0;}