字符串hash

最新推荐文章于 2022-05-01 14:12:17 发布

原创最新推荐文章于 2022-05-01 14:12:17 发布 · 180 阅读

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选用 BKDRhash 方式
参考资料：https://www.byvoid.com/blog/string-hash-compare/
字符串通篇资料：https://blog.youkuaiyun.com/ck_boss/article/details/47066727

计算拼接字符串哈希值：

//就这一破式子我推了半天
hash(a) * pow(p, length(bc)) + hash(bc)

HDU 4821 string

传送门：https://vjudge.net/problem/HDU-4821
题意：给出M和L，和一个字符串S。要求找出S的子串中长度为L*M，并且可以分成M段，每段长L，并且M段都不相同的子串个数。

思路：一道字符串哈希题。哈希的方法是BKDRHash，哈希中进制是31,131等素数。

	#include <iostream>
#include <map>
#include <string>
#include <cstring>
#include <cstdio>
#define maxn 100005
#define SEED 31
#define ULL unsigned long long
using namespace std;

ULL k[maxn],ha[maxn];
map <ULL , int> mp;
int m,l;
char s[maxn];
int main(){
	while (scanf("%d%d",&m,&l)!=EOF){
		scanf("%s",s);
		k[0]=1;
		int len = strlen(s);
		ha[len] = 0;
		for (int i=1;i<=l;i++)
			k[i] = k[i-1]*SEED;
		for (int i=len-1;i>=0;i--)
			ha[i] = ha[i+1]*SEED + s[i] - 'a' + 1;
		int ans=0;

		for (int i=0;i<l && i+m*l<len;i++){
			mp.clear();
			for (int j=i;j<i+m*l;j+=l)
				mp[ha[j]-ha[j+l]*k[l]]++;
			if (mp.size() == m) ans++;
			for (int j=i+m*l;j<=len-l;j+=l){
				int head = j-m*l;
				ULL x = ha[head] - ha[head+l]*k[l];
				mp[x]--;
				if (mp[x] == 0) mp.erase(x);
				mp[ha[j] - ha[j+l]*k[l]]++;
				if (mp.size() == m) ans ++;
			}
		}
		cout << ans << endl;
	}
}

HDU 4080 Stammering Aliens
题意
每组数据为一个整数m和一个长度不小于m的字符串，求该字符串的一个子串，该子串在满足出现次数不小于m的同时应尽量长。
输出该长度和最右侧出现的起始位置。如果存在多组数据，输出有最靠近右侧的那组。
m<=4e5,时限5s

字符串hash+二分答案。我写的版本被卡了，从网上找了一个相同写法的4000ms卡过去了，学习一下。
by Vicente:

#include <cstring>
#include <iostream>
#include <map>
#define maxn 40005
#define SEED 31
#define ULL unsigned long long 
using namespace std;
int m,len;
ULL k[maxn],ha[maxn];
char s[maxn];
map <ULL , int> mp;

bool isOK(int L){
	mp.clear();
	for (int i=0;i+L<=len;i++){
		mp[ha[i]-ha[i+L]*k[L]]++;
		if (mp[ha[i] - ha[i+L]*k[L]]>=m)return true; 
	}
	return false;
}
int main(){
	while (scanf("%d",&m) &&  m!=0){
		scanf("%s",s);
		len = strlen(s);
		k[0] = 1;
		for (int i=1;i<=len;i++) k[i] = k[i-1] * SEED;
		ha[len] = 0;
		for (int i = len-1;i>=0;i--) ha[i] = ha[i+1] * SEED + s[i] - 'a'+1;
		int l=m,r=len,mid;
		while (l<r){
			mid = (l+r+1) >>1;
			if (isOK(mid)) l = mid;else r=mid-1;
		}

		if (!isOK(r)) {
			puts("none");
			continue;
		}else{
			cout << r << " ";
			mp.clear();
			for (int i=0;i+r<=len;i++){
				mp[ha[i]-ha[i+r]*k[r]]++;
				if (mp[ha[i] - ha[i+r]*k[r]]>=m) cout << i << endl; 
			}
		}
	}
}

AC版本：

#include <cstring>
#include <iostream>
#include <map>
#define ull unsigned long long
using namespace std;
const int maxn = (int)4e4 + 10;
int n,len;
char s[maxn];
ull has[maxn];
ull p = 2333,a[maxn];
 
bool judge(int x)
{
	map<int,int> mp;
	ull res;
	for (int i = 1;i + x - 1 <= len;i ++)
	{
		res = has[i + x - 1] - has[i - 1] * a[x];
		if (++mp[res] >= n) return 1;
	}
	return 0;
}
int check(int x)
{
	map<int,int> mp;
	int pos = -1;
	ull res;
	for (int i = 1;i + x - 1 <= len;i ++)
	{
		res = has[i + x - 1] - has[i - 1] * a[x];
		if (++mp[res] >= n) pos = i;
	}
	return pos - 1;
}
int main()
{
	while (~scanf("%d",&n) && n)
	{
		scanf("%s",s + 1);
		len = strlen(s + 1);
		has[0] = 0,a[0] = 1;
		for (int i = 1;i <= len;i ++)
		{
			has[i] = has[i - 1] * p + s[i];
			a[i] = a[i - 1] * p;
		}
		int l = 1,r = len + 1,mid;
		while (l <= r)
		{
			mid = (l + r) >> 1;
			if (judge(mid)) l = mid + 1;
			else r = mid - 1;
		}
		if (r == 0)
			printf("none\n");
		else
			printf("%d %d\n",r,check(r));
	}
	return 0;	
}

哈哈哈经过学习，我改了一点内容，终于也把这道题卡过去了。
在这里插入图片描述
4773ms…说是卡过去真是一点不错。
AC代码 by Vicente.

#include <cstring>
#include <iostream>
#include <map>
#define maxn 40005
#define SEED 31
#define ULL unsigned long long 
using namespace std;
int m,len;
ULL k[maxn],ha[maxn];
char s[maxn];
map <ULL , int> mp;

bool isOK(int L){
	mp.clear();
	ULL res;
	for (int i=0;i+L<=len;i++){
		res=ha[i]-ha[i+L]*k[L];
		
		if (++mp[res]>=m)return true; 
	}
	return false;
}

int main(){
	while (scanf("%d",&m) &&  m!=0){
		scanf("%s",s);
		len = strlen(s);
		k[0] = 1;
		ha[len] = 0;
		for (int i=1;i<=len;i++)//强行放在一起写，就没爆T了
			k[i] = k[i-1] * SEED,ha[len-i] = ha[len-i+1] * SEED + s[len-i] - 'a'+1;;
		// for (int i = len-1;i>=0;i--) ha[i] = ha[i+1] * SEED + s[i] - 'a'+1;
		int l=1,r=len+1,mid;

		while (l<=r){
			mid = (l+r) >>1;
			if (isOK(mid)) l = mid+1;else r=mid-1;
		}

		if (r == 0) {
			puts("none");
			continue;
		}else{
			cout << r << " ";
			mp.clear();
			int x;
			for (int i=0;i+r<=len;i++){
				mp[ha[i]-ha[i+r]*k[r]]++;
				if (mp[ha[i] - ha[i+r]*k[r]]>=m) x=i;//cout << i << endl; 
			}
			cout << x << endl;//不用x标记的话会WA，原因自己想
		}
	}
}