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题意:训练之南P225
分析:二分寻找长度,用hash值来比较长度为L的字串是否相等。
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
const int N = 4e4 + 5;
const int x = 123;
ull H[N], _hash[N], xp[N];
int rk[N];
char str[N];
int m;
void get_hash(char *s, int len) {
H[len] = 0;
for (int i=len-1; i>=0; --i) {
H[i] = H[i+1] * x + (s[i] - ‘a‘);
}
xp[0] = 1;
for (int i=1; i<len; ++i) {
xp[i] = xp[i-1] * x;
}
}
bool cmp(const int &a, const int &b) {
return (_hash[a] < _hash[b] || (_hash[a] == _hash[b] && a < b));
}
int check(int L, int len) {
int cnt = 0, pos = -1, c = 0;
for (int i=0; i<len-L+1; ++i) {
rk[i] = i;
_hash[i] = H[i] - H[i+L] * xp[L];
}
sort (rk, rk+len-L+1, cmp);
for (int i=0; i<len-L+1; ++i) {
if (i == 0 || _hash[rk[i]] != _hash[rk[i-1]]) c = 0;
if (++c >= m) pos = max (pos, rk[i]);
}
return pos;
}
int main(void) {
while (scanf ("%d", &m) == 1) {
if (!m) break;
scanf ("%s", &str);
int len = strlen (str);
get_hash (str, len);
if (check (1, len) == -1) puts ("none");
else {
int l = 1, r = len + 1;
while (r - l > 1) {
int mid = l + r >> 1;
if (check (mid, len) >= 0) l = mid;
else r = mid;
}
printf ("%d %d\n", l, check (l, len));
}
}
return 0;
}
字符串hash LA 4513 Stammering Aliens
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原文地址:http://www.cnblogs.com/Running-Time/p/5123779.html