HDU 1711 Number Sequence (KMP簡單題)
阿新 • • 發佈:2018-08-11
bmi AMM several rip case ive sts -- 沒有
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.
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分析:
題目意思:
給你一個長的文本串,一個短的模板串
問你模板串匹在文本串中匹配到的位置
沒有匹配到的話,輸出-1
跑kmp就好
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39408 Accepted Submission(s): 16269
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
#include<cstdio> #include<iostream> #include<cstring> #include<memory> using namespace std; int moban[1000005],wenben[1000005]; int next1[1000005]; int sum; void getnext(int* s,int* next1,int m) { next1[0]=0; next1[1]=0; for(int i=1;i<m;i++) { int j=next1[i]; while(j&&s[i]!=s[j]) j=next1[j]; if(s[i]==s[j]) next1[i+1]=j+1; else next1[i+1]=0; } } void kmp(int* ss,int* s,int* next1,int n,int m) { int ans=-1; getnext(s,next1,m); int j=0; for(int i=0;i<n;i++) { while(j&&s[j]!=ss[i]) j=next1[j]; if(s[j]==ss[i]) j++; if(j==m) { ans=i-m+2; break; } } printf("%d\n",ans); } int main() { int t; scanf("%d",&t); int n,m; while(t--) { scanf("%d %d",&n,&m); for(int i=0;i<n;i++) scanf("%d",&wenben[i]); for(int i=0;i<m;i++) scanf("%d",&moban[i]); kmp(wenben,moban,next1,n,m); } return 0; }
HDU 1711 Number Sequence (KMP簡單題)