Number Sequence(HDU1711)

[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher

题目：

A - Number Sequence

Time Limit:5000MS

Memory Limit:32768KB

HDU 1711

Description

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.

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
. 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

题目大意：

两个数组a,b，a作为主串，b为模板串，寻找a中包含b的最小子串下标

分析：

法一 a作为主串，b作为模板串，用kmp匹配

（我学了好久kmp才明白这个算法） 我是看了俞勇的《ACM国际大学生程序设计竞赛知识与入门》p161懂了匹配的方式

看了http://www.cnblogs.com/c-cloud/p/3224788.html才明白next数组怎么构造

尤其是这两张图特别容易让人明白next数组构造方式





kmp模板：

//p[]为模板串，n为模板串的长度，现构造next数组
void makenext(int n)
{
next[0] = 0;
for (int i = 1, j = 0; i < n; i++)
{
while (j>0 && p[j] != p[i])
j = next[j - 1];
if (p[i] == p[j])
{
j++;
}
next[i] = j;
}
}

//T[]为主串，m为主串的长度，p[]为模板串，n为模板串的长度
int kmp(int m, int n)
{
for (int i = 0, j = 0; i < m; i++)
{
while (j>0 && T[i] != p[j])
j = next[j - 1];
if (T[i] == p[j])
{
j++;
}
if (j == n)return i - n + 1;
}
return -1;//主串并不包含模板串
}

法二 用hash字符串处理，匹配主串和模板串

看的是大白书（《挑战程序设计竞赛》）p374

只要不断这样计算开始位置右移一位后的字符串子串的哈希值，就可以在O(n)时间内得到所有位置对应的hash值，从而在O(n+m)时间内完成字符串匹配。在实现时，可以用64位无符号整数计算hash值，并取h为2^64，通过自然溢出省去求模运算。

hash模板：

//al为主串长度，bl为模板串长度

typedef unsigned long long ull;
const ull B=100000007;//hash基数

int contain(int al, int bl)
{
if (bl > al)return -1;
int cnt = 0;
//计算B的bl次方
ull t = 1;
for (int i = 0; i < bl; i++)
t *= B;

//计算a和b长度为bl的前缀对应的hash值
ull ah = 0, bh = 0;
for (int i = 0; i < bl; i++)ah = ah*B + a[i];
for (int i = 0; i < bl; i++)bh = bh*B + b[i];

//对a不断右移一位，更新hash值并判断
for (int i = 0; i+bl <=al; i++)
{
if (ah == bh)return i;
if(i+bl<al)
ah = ah*B - a[i] * t + a[i + bl];
}
return -1;
}

用kmpAC代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<string.h>
using namespace std;
#define maxnA 1000001
#define maxnB 10001
int a[maxnA];
int b[maxnB];
int nextb[maxnB];
void makenext(int bl)
{
nextb[0] = 0;
for (int i = 1, j = 0; i < bl; i++)
{
while (j>0 && b[i] != b[j])
j = nextb[j - 1];
if (b[i] == b[j])
{
j++;
}
nextb[i] = j;
}
}

int kmp(int al,int bl)
{
makenext(bl);
for (int i = 0, j = 0; i < al; i++)
{
while (j > 0 && b[j] != a[i])
j = nextb[j - 1];
if (b[j] == a[i])
j++;
if (j == bl)
return i - bl + 1;
}
return -2;
}

int main()
{
int cas;
scanf("%d", &cas);
while (cas--)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(nextb, 0, sizeof(nextb));
int al, bl;
scanf("%d%d", &al, &bl);
for (int i = 0; i < al; i++)
scanf("%d", &a[i]);
for (int i = 0; i < bl; i++)
scanf("%d", &b[i]);
int res = kmp(al, bl)+1;
printf("%d\n", res);
}
}

用hash字符串AC代码：

#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
typedef unsigned long long ull;
const ull B = 100000007;
const int MAXN = 1000001;
const int MAXM = 10001;
int a[MAXN];
int b[MAXM];

//b是否在a中
int contain(int al, int bl)
{
if (bl > al)return -2;
int cnt = 0;
//计算B的bl次方
ull t = 1;
for (int i = 0; i < bl; i++)
t *= B;

//计算a和b长度为bl的前缀对应的hash值
ull ah = 0, bh = 0;
for (int i = 0; i < bl; i++)ah = ah*B + a[i];
for (int i = 0; i < bl; i++)bh = bh*B + b[i];

//对a不断右移一位，更新hash值并判断
for (int i = 0; i < al; i++)
{
if (ah == bh)return i;
else
ah = ah*B - a[i] * t + a[i + bl];
}
return -2;
}

int main()
{
int cas;
scanf("%d", &cas);
while (cas--)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
int al, bl;
scanf("%d%d", &al, &bl);
for (int i = 0; i < al; i++)
scanf("%d", &a[i]);
for (int i = 0; i < bl; i++)
scanf("%d", &b[i]);
int res = contain(al, bl)+1;
printf("%d\n", res);
}
}

小结：

个人觉得大多数kmp匹配问题似乎都可以用hash处理，而且hash似乎好写一些（较易理解），况且hash比kmp省了300ms。