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UVA 11107 - Life Forms

题目链接

题意：给定一个字符串，找出重复出现超过m次的字串的最大开始下标

思路：hash大法，需要点人品，然后二分答案，每次利用hash值去找出最大下标即可

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef unsigned long long ull;

const ull x = 123;
const int N = 40005;

int m, n;
char str[N];
ull H[N], Hp[N];

void gethash() {
    H[n] = 0;
    for (int i = n - 1; i >= 0; i--)
	H[i] = H[i + 1] * x + str[i] - 'a';
}

struct Suf {
    int hash, id;
} suf[N];

bool cmp(Suf a, Suf b) {
    if (a.hash == b.hash) return a.id < b.id;
    return a.hash < b.hash;
}

int find(int L) {
    for (int i = 0; i < n - L + 1; i++) {
	suf[i].hash = H[i] - H[i + L] * Hp[L];
	suf[i].id = i;
    }
    sort(suf, suf + n - L + 1, cmp);
    int ans = -1;
    for (int i = 0; i < n - L + 1; i++) {
	int j, len = 0;
	for (j = i; suf[i].hash == suf[j].hash && j < n - L + 1; j++)
	    len++;
	if (len >= m) {
	    ans = max(ans, suf[j - 1].id);
	}
	i = j - 1;
    }
    return ans;
}

void solve() {
    if (find(1) == -1) printf("none\n");
    else {
	int l = 1, r = n - m + 2;
	while (l < r) {
	    int mid = (l + r) / 2;
	    if (find(mid) == -1) r = mid;
	    else l = mid + 1;
	}
	printf("%d %d\n", l - 1, find(l - 1));
    }
}

int main() {
    Hp[0] = 1;
    for (int i = 1; i < N; i++)
	Hp[i] = Hp[i - 1] * x;
    while (~scanf("%d", &m) && m) {
	scanf("%s", str);
	n = strlen(str);
	gethash();
	solve();
    }
    return 0;
}

UVA 11107 - Life Forms(Hash+LCP),布布扣,bubuko.com

UVA 11107 - Life Forms(Hash+LCP)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38639955