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hdu4821 string (字符串Hash)

2016-03-06 21:39 387 查看

参考了网上的代码；

这题主要用的就是Rabin-Karp的hash算法原理；

dp[i]表示i后面字符串的hash值，要算出S[i...i+m]的hash值的话就是：tmp = (dp[j] - dp[j+len]*nbase[len])；

这题问的就是原串有多少个满足条件的substring；

(i) It is of length M*L;

(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings
are considered as “diversified” if they don’t have the same character for every position.

这里只需要枚举每个串的头就行了，substring中含有n个长为len的串，那么枚举了头i，然后求出后面连续n个长为len 的串的hash值放到mp中，满足条件的充要条件就是mp的大小为n，然后把前面的第一个长为len的hash值-1，如果为0 的话就移除mp中，然后后面又加入一个长为len 的串的hash值，具体操作看代码。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :Rabin-karp hash Algorithm.
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e5 + 10;
char s[maxn];
int n, len, slen;
ULL nbase[maxn];
ULL dp[maxn];
const ULL base = 31;
map<ULL,int> mp;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
nbase[0] = 1ULL;
int i, j;
for (i = 1;i < maxn;++i)
nbase[i] = nbase[i-1]*base;
while(~scanf("%d%d",&n,&len)) {
scanf("%s", s);
// puts(s);
slen = strlen(s);
dp[slen] = 0;
for (i = slen - 1;i >= 0;--i) {
dp[i] = dp[i+1]*base + (s[i] - '0');
// printf("*******   %lld\n", dp[i]);
}
int answer = 0;
for (i = 0;i < len && i + n*len <= slen;++i) {
mp.clear();
for (j = i;j < i + n * len;j += len) {
ULL tmp = (dp[j] - dp[j+len]*nbase[len]);
mp[tmp]++;
// printf("%lld\n",tmp);
}
if (mp.size() == n) answer++;
for (j = i+n*len;j + len <= slen;j += len) {
ULL tmp = dp[j-n*len] - dp[j-(n-1)*len]*nbase[len];
mp[tmp]--;
if (mp[tmp] == 0) mp.erase(tmp);
tmp = dp[j] - dp[j+len]*nbase[len];
mp[tmp]++;
if (mp.size() == n) answer++;
}
}
printf("%d\n", answer);
}
return 0;
}

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