HDU 1711 Number Sequence 【KMP应用 求成功匹配子串的最小下标】

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42917    Accepted Submission(s): 17715

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

题意概括：

给出一串主串，一串子串；

求成功匹配到子串的最小下标；

解题思路：

KMP的应用，稍微变形，匹配到子串就跳出来输出。

AC code：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 1e6+10;
const int MAXM = 1e4+10;
int W[MAXM], T[MAXN];
int wlen, tlen;
int nxt[MAXM];

void get_nxt()
{
    int j, k;
    j = 0;
    k = -1;
    nxt[0] = -1;
    while(j < wlen){
        if(k == -1 || W[j] == W[k]){
            nxt[++j] = ++k;
        }
        else k = nxt[k];
    }
}

int KMP_index()
{
    int i = 0, j = 0;
    get_nxt();

    while( i < tlen && j < wlen){
        if(j == -1 || T[i] == W[j]){
            i++;
            j++;
        }
        else j = nxt[j];
    }
    if(j == wlen) return i-wlen+1;
    else return -1;
}

int main()
{
    int T_case, N, M;
    scanf("%d", &T_case);
    while(T_case--)
    {
        scanf("%d%d", &N, &M);
        wlen = M, tlen = N;
        for(int i = 0; i < N; i++)
            scanf("%d", &T[i]);
        for(int j = 0; j < M; j++)
            scanf("%d", &W[j]);

        printf("%d
", KMP_index());
    }
    return 0;
}