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HDU 1711 Number Sequence（kmp）
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
还是kmp，就是把字符变成整型数组了，还是模板
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
#define N 1000005
int n,m;
int a[N],b[N];
int next[N];
void getfail(int *b)
{
int i,j;
next[0]=-1;
j=-1;
i=0;
while(i<m)
if(j==-1||b[i]==b[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
int kmp(int *a,int *b)
{
int i,j;
i=j=0;
while(i<n)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else
j=next[j];
if(j==m)
return i-j+1;
}
return -1;
}
int main()
{
int t,i,j;
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]);
getfail(b);
printf("%d\n",kmp(a,b));
}
return 0;
}
HDU 1711 Number Sequence（kmp）
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