hdu 1711 Number Sequence kmp模式匹配

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30458    Accepted Submission(s): 12830

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest 

#include<stdio.h>#include<algorithm>using namespace std;#define  size 10010int a[1000010], b[size];int nextval[size];int n, m;void getNextval(int *T) {    int k = -1;    int j = 0;    nextval[j] = k;    while (j  < m) {        if ((k == -1) || (T[j] == T[k])) {            ++k;            ++j;            nextval[j] = k;        }        else {            k = nextval[k];        }    }}//S为主串  从主串pos位置开始匹配int index_KMP(int *S, int *T, int pos) {    int i;    int j;    i = pos;    j = 0;    while ((i < n) && (j < m)) {        /* j = -1 表示next[0], 说明失配处在模式串T的第0个字符。所以这里特殊处理，然后令i+1和j+1。*/        if ((j == -1) || S[i] == T[j]) {            i++;            j++;        }        else {            j = nextval[j];        }    }    //找到    if (m == j)        return i - m;    else        return  -2;}int main() {    int test;    scanf("%d", &test);    while (test--) {        scanf("%d%d", &n, &m);        for (int i = 0; i < n; i++) {            scanf("%d",&a[i]);        }        for (int i = 0; i < m; i++) {            scanf("%d",&b[i]);        }        getNextval(b);        int k = index_KMP(a, b, 0) + 1;        printf("%d\n", k);    }}