标签：kmp 模板
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11817 Accepted Submission(s): 5395
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
KMP模板
代码：
#include <stdio.h>
#include <string.h>
#define M 10005
#define N 1000005
int a[N], b[M], next[M], n, m;
void get_next(){
int i, j, k;
i = 1; j = 0;
while(i<= m){
if(j==0||b[i] == b[j]){
++i; ++j; next[i] = j;
}
else j = next[j];
}
}
int kmp(){
int i, j;
i = 1, j = 1;
while(i<=n&&j<= m){
if(j == 0||a[i] == b[j]){
++i; ++j;
}
else j = next[j];
}
if(j > m) return i-j+1;
return -1;
}
int main(){
int t;
scanf("%d", &t);
while(t --){
scanf("%d%d", &n, &m);
memset(next, 0, sizeof(next));
next[1] = 0;
int i, j;
for(i = 1; i <= n; i ++) scanf("%d", &a[i]);
for(i = 1; i <= m; ++ i) scanf("%d", &b[i]);
get_next();
int ans = kmp();
printf("%d\n", ans);
}
return 0;
}
hdoj 1711 Number Sequence 【KMP】
标签：kmp 模板
原文地址：http://blog.csdn.net/shengweisong/article/details/40904395
