HDOJ1711 KMP模板程序

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29918 Accepted Submission(s): 12609

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

Recommend
lcy | We have carefully selected several similar problems for you: 1358 1686 3746 2222 1867

还是当做字符串匹配的问题计算
预处理的next数组好像有两种方法，一个是next【0】=-1，一个是next【0】=0；
大同小异吧，我的是next【0】=0。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const int maxn = 1e6+5;
const int maxm = 1e4+5;
int i,j,k,n,m;
int a[maxn],b[maxm],nex[maxm];


void make_next(){
    nex[0] = nex[1] = 0;
    for (i=1; i<m; i++) {
        k=nex[i];
        while (k && b[i]!=b[k]) k = nex[k];
        nex[i+1] = b[i]==b[k]?k+1:0;
    }
}

int kmp(){
   k =0;
   for (i=0; i<n; i++) {
      while (k && a[i]!=b[k]) k = nex[k];
      if (a[i]==b[k]) k++;
      if (k==m) return i-m+2;
   }
   return -1;
}

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