hdu1711 KMP应用之详细讲解
Kmp算法 详细 看严蔚敏的视频教程 很详细 很好
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5047 Accepted Submission(s): 2275
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
题意 输入2串数字
第二串数在第一串数中出现的位置
#include<stdio.h>int a[1000005];int b[10005];int next[10005];int m,n;void get_next(){ int i,j; next[1]=0; j=0;i=1; while(i<m)//不要等于m { if(j==0||b[i]==b[j]) {i++;j++;next[i]=j;}//这个地方完全可以弄到m所以等于的时候不用继续循环 else j=next[j]; }}int KMP()//注意此函数和get_next可不一样{ int i,j; next[1]=0; i=1;j=1;//这里ij分别是二者的起始位置 while(i<=n&&j<=m) { if(j==0||a[i]==b[j]) {i++;j++;}//如果相等则比较下一个 不等就回溯 这里不要再给next数组赋值了上面已经赋好了 else j=next[j]; } if(j>m) return i-m; else return -1;}int main(){ int i,t,ans; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&m); for(i=1;i<=n;i++) scanf("%d",&a[i]);//注意从1开始输入 for(i=1;i<=m;i++) scanf("%d",&b[i]); get_next(); ans=KMP(); printf("%d\n",ans); } return 0;}