HDU1711:Number Sequence
2016-03-20 16:34
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[align=left]Problem Description[/align]
Given two sequences of numbers : a[1], a[2], ...... , a
, 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.
[align=left]Input[/align]
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
. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
6
-1
//low方法应该是一点一点的找,但是,运用KMP算法会变牛逼。。(KMP竟然还可以用到数组中!!(⊙o⊙)
//这题很有趣,让我知道算法更重要的是思想,不要有局限自己思维
Given two sequences of numbers : a[1], a[2], ...... , a
, 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.
[align=left]Input[/align]
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
. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
6
-1
//low方法应该是一点一点的找,但是,运用KMP算法会变牛逼。。(KMP竟然还可以用到数组中!!(⊙o⊙)
//这题很有趣,让我知道算法更重要的是思想,不要有局限自己思维
#include <iostream> #include <cstdio> using namespace std; int data1[1000003],data2[1000003]; int nex[1000003],n,m; void getnext() { int i=0,j=-1; nex[0]=-1; while(i<m) { if(j==-1||data2[i]==data2[j]) { i++;j++; if(data2[i]==data2[j]) nex[i]=nex[j]; else nex[i]=j; } else j=nex[j]; } } int kmp() { int i=0,j=0; getnext(); while(i<n) { if(j==-1||data1[i]==data2[j]) { i++;j++; } else { j=nex[j];// j=nex[i]这里又记错了。。。。 } if(j==m) return i-m+1; } return -1; } int main() { int l; cin>>l; while(l--) { cin>>n>>m; for(int i=0;i<n;i++) scanf("%d",&data1[i]); for(int i=0;i<m;i++) scanf("%d",&data2[i]); getnext(); int ans=kmp(); cout<<ans<<endl; } return 0; }
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