Cow Patterns poj 3167 (hash解法)

围观了叉姐的代码，hash对于解串类题目真的好灵活， 又涨姿势了 ==。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>   
#include <map>
#include <string>  
#include <climits> 
#include <set>
#include <string>    
#include <sstream>
#include <utility>   
#include <ctime>
#include <bitset>

using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::stringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
using std::unique;
using std::lower_bound;
using std::random_shuffle;
using std::bitset;
using std::upper_bound;
using std::multiset;

typedef long long LL;
typedef unsigned long long ULL;
typedef unsigned UN;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
typedef LL TY;
typedef long double LF;

const int MAXN(100010);
const int MAXM(55);
const int MAXE(200010);
const int MAXK(6);
const int HSIZE(13131);
const int SIGMA_SIZE(26);
const int MAXH(20);
const int INFI((INT_MAX-1) >> 1);
const ULL BASE(31);
const LL LIM(1e13);
const int INV(-10000);
const int MOD((1 << 30)-1);
const double EPS(1e-7);
const LF PI(acos(-1.0));

template<typename T> inline void checkmax(T &a, T b){if(b > a) a = b;}
template<typename T> inline void checkmin(T &a, T b){if(b < a) a = b;}
template<typename T> inline T ABS(const T &a){return a < 0? -a: a;}

ULL hash1[30], hash2[30];
int arr1[100010], arr2[25010];
int ans[100010], ca;

int main()
{
	int n, K, s;
	while(~scanf("%d%d%d", &n, &K, &s))
	{
		for(int i = 1; i <= n; ++i) scanf("%d", arr1+i);
		for(int i = 1; i <= K; ++i) scanf("%d", arr2+i);
		if(n < K)
		{
			printf("0\n");
			continue;
		}
		ULL sub = 1;
		for(int i = 1; i < K; ++i) sub *= BASE;
		memset(hash2, 0, sizeof(hash2));
		for(int i = 1; i <= K; ++i)
		{
			for(int j = 0; j < s; ++j) hash2[j] *= BASE;
			++hash2[arr2[i]-1];
		}
		memset(hash1, 0, sizeof(hash1));
		for(int i = 1; i < K; ++i)
		{
			for(int j = 0; j < s; ++j) hash1[j] *= BASE;
			++hash1[arr1[i]-1];
		}
		ca = 0;
		for(int i = K; i <= n; ++i)
		{
			for(int j = 0; j < s; ++j) hash1[j] *= BASE;
			++hash1[arr1[i]-1];
			int t1 = s-1, t2 = s-1;
			while(true)
			{
				while(t1 >= 0 && hash1[t1] == 0) --t1;
				while(t2 >= 0 && hash2[t2] == 0) --t2;
				if(t1 == -1 && t2 == -1) break;
				if(t1 == -1 || t2 == -1) break;
				if(hash1[t1] != hash2[t2]) break;
				--t1;
				--t2;
			}
			if(t1 == -1 && t2 == -1) ans[ca++] = i-K+1;
			hash1[arr1[i-K+1]-1] -= sub;
		}
		printf("%d\n", ca);
		for(int i = 0; i < ca; ++i) printf("%d\n", ans[i]);
	}
	return 0;
}