hdu 1711 Number Sequence KMP

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

分析：求最小偏移位置使得两字符串匹配，KMP应用。

/*Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12466    Accepted Submission(s): 5687


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
 

Source
HDU 2007-Spring Programming Contest
*/
#include <cstdio>
const int maxn = 10000 + 10, maxm = 1000000 + 10;
int n, m;
int p[maxn], t[maxm], next[maxn]; //定义模式串，文本串，next数组 
void getNext() //O(m)复杂度求Next数组 
{
    int i = 0, j = 0, k = -1;
    next[0] = -1;
    while(j < m){
        if(k == -1 || p[k] == p[j]) next[++j] = ++k;
        else k = next[k];
    }
}
int kmp()
{
    int j = 0; //初始化模式串的前一个位置 
    getNext(); //生成next数组 
    for(int i = 0; i < n; i++){//遍历文本串 
        while(j && p[j] != t[i]) j = next[j];//持续走直到可以匹配 
        if(p[j] == t[i]) j++; //匹配成功继续下一个位置 
        if(j == m) return i-m+2; //找到后返回第一个匹配的位置 
    }
    return -1; //找不到返回-1 
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++) scanf("%d", t+i);
        for(int i = 0; i < m; i++) scanf("%d", p+i);
        printf("%d\n", kmp());
    }
    return 0;
}