604
HDU1711-Number Sequence
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11038 Accepted Submission(s): 5038
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
字符串匹配题
KMP算法
AC代码：
#include<stdio.h>
#include<string.h>
int a[1000010],b[10010],next[10010];
int n,m;
void getnext(int p[])
{
int i = 0;
int j = -1;
next[0] = -1;
while(i < m)
{
if(j == -1 || p[i] == p[j])
{
i++;
j++;
if(p[i]!=p[j])
next[i] = j;
else
next[i] = next[j];
}
else
j = next[j];
}
}
int KMP(int a[],int b[])
{
int j = 0;
int i = 0;
getnext(b);
while(i < n && j < m)
{
if(j == -1 || a[i] == b[j])
{
i++;
j++;
}
else
j = next[j];
}
if(j >= m)
return i - m+1;
else
return -1;
}
int main()
{
int i,j,T;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(next,0,sizeof(next));
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&b[i]);
printf("%d\n",KMP(a,b));
}
return 0;
}
最后更新：2017-04-03 05:39:50