uva 12206 - Stammering Aliens(雜湊)

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題目連結：uva 12206 - Stammering Aliens

題目大意：給出一個字串，找出至少出現m次的最長子串。

解題思路：雜湊演算法，將每個尾碼數組建立一個雜湊值，每次二分長度判斷，每次判斷時將雜湊值排序，計數即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long ll;const int maxn = 40005;const int x = 123;int N, M, pos, Rank[maxn];char str[maxn];ll h[maxn], xp[maxn], Hash[maxn];void init () {    scanf("%s", str);    N = strlen(str);    h[N] = 0;    for (int i = N - 1; i >= 0; i--)        h[i] = h[i+1] * x + str[i] - ‘a‘;    xp[0] = 1;    for (int i = 1; i <= N; 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);}bool judge (int l) {    int c = 0, n = N - l + 1;    pos = -1;    for (int i = 0; i < n; i++) {        Rank[i] = i;        Hash[i] = h[i] - h[i+l] * xp[l];    }    sort (Rank, Rank + n, cmp);    for (int i = 0; i < n; i++) {        if (i == 0 || Hash[Rank[i]] != Hash[Rank[i-1]]) c = 0;        if (++c >= M)            pos = max(pos, Rank[i]);    }    return pos >= 0;}void bsearch () {    if (!judge(1)) {        printf("none\n");        return;    }    int l = 1, r = N + 1;    while (r - l > 1) {        int mid = (r + l) / 2;        if (judge(mid))            l = mid;        else            r = mid;    }    judge(l);    printf("%d %d\n", l, pos);}int main () {    while (scanf("%d", &M) == 1 && M) {        init();        bsearch();    }    return 0;}

uva 12206 - Stammering Aliens(雜湊)