HDOJ 5442 Favorite Donut

最新推荐文章于 2019-01-27 14:31:43 发布

Lost_in_wine

最新推荐文章于 2019-01-27 14:31:43 发布

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分类专栏：后缀数组 KMP 最小表示法

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后缀数组

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最小表示法

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Favorite Donut

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1276 Accepted Submission(s): 336

Problem Description

Lulu has a sweet tooth. Her favorite food is ring donut. Everyday she buys a ring donut from the same bakery. A ring donut is consists of

n parts. Every part has its own sugariness that can be expressed by a letter from

a to

z (from low to high), and a ring donut can be expressed by a string whose i-th character represents the sugariness of the

i−th part in clockwise order. Note that

z is the sweetest, and two parts are equally sweet if they have the same sugariness.

Once Lulu eats a part of the donut, she must continue to eat its uneaten adjacent part until all parts are eaten. Therefore, she has to eat either clockwise or counter-clockwise after her first bite, and there are

2n ways to eat the ring donut of

n parts. For example, Lulu has

6 ways to eat a ring donut

abc :

abc,bca,cab,acb,bac,cba . Lulu likes eating the sweetest part first, so she actually prefer the way of the greatest lexicographic order. If there are two or more lexicographic maxima, then she will prefer the way whose starting part has the minimum index in clockwise order. If two ways start at the same part, then she will prefer eating the donut in clockwise order. Please compute the way to eat the donut she likes most.

Input

First line contain one integer

T,T≤20 , which means the number of test case.

For each test case, the first line contains one integer

n,n≤20000 , which represents how many parts the ring donut has. The next line contains a string consisted of

n lowercase alphabets representing the ring donut.

Output

You should print one line for each test case, consisted of two integers, which represents the starting point (from

1 to

n ) and the direction (

0 for clockwise and

1 for counterclockwise).

Sample Input

  
   2
4
abab
4
aaab

Sample Output

  
   2 0
4 0

Source

2015 ACM/ICPC Asia Regional Changchun Online

最小表示法 +KMP

#include <cstdio>
#include <cstring>

const int N = 20010;

int get_min(char *str, int len) {
	int i, j, k;
	i = k = 0, j = 1;
	while (i < len && j < len && k < len) {
		int t = str[(i + k) % len] - str[(j + k) % len];
		if (t == 0)
			++k;
		else {
			if (t < 0) i += k + 1;
			else j += k + 1;
			if (i == j) ++j;
			k = 0;
		}
	}
	return i < j ? i : j;
}

void change(char* S1, char* S2, int len) {
	for (int i = 0; i <= len; ++i)
		S2[i] = S1[len - i - 1];
}

void get_next(char *str, int *next) {
	int i = 0, j = -1;
	next[i] = j;

	while (str[i]) {
		if (j == -1 || str[i] == str[j])
			next[++i] = ++j;
		else
			j = next[j];
	}
}

int main() {
	int T, len;
	char S1[N], S2[N];
	int next[N];

	scanf("%d", &T);
	while (T--) {
		scanf("%d%s", &len, S1);
		change(S1, S2, len);

		int x = get_min(S1, len);
		int y = get_min(S2, len);
		get_next(S2, next);

		int t = len - next[len];
		if (t && t != len && (len % t == 0)) {
			y += len - t;
		}

		int flag = 0;
		int k;
		for (k = 0; k < len; ++k) {
			if (S1[(x + k) % len] > S2[(y + k) % len])
				break;
			else if (S1[(x + k) % len] < S2[(y + k) % len]) {
				flag = 1;
				break;
			}
		}
		y = len - y;
		++x;
		if (k < len) {
			if (flag == 1)
				printf("%d 1\n", y);
			else
				printf("%d 0\n", x);

		}
		else {
			if (x <= y)
				printf("%d 0\n", x);
			else
				printf("%d 1\n", y);
		}
	}

	return 0;
}

后缀数组

#include <cstdio>
#include <cstring>

const int N = 51000;

int wa[N], wb[N], wv[N], ws[N];
char S1[N], S2[N], TT[N];
int sa[N];
int len;

bool Cmp(int *r, int a, int b, int l) {
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void DA(char* ary, int n, int m) {
    int i, j, p, *x = wa, *y = wb, *t;
    for (i = 0; i < m; ++i) ws[i] = 0;
    for (i = 0; i < n; ++i) ws[x[i] = ary[i]]++;
    for (i = 1; i < m; ++i) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; --i) sa[--ws[x[i]]] = i;

    for (j = p = 1; p < n; m = p, j *= 2) {
        for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
        
        for (i = 0; i < n; ++i) wv[i] = x[y[i]];
        for (i = 0; i < m; ++i) ws[i] = 0;
        for (i = 0; i < n; ++i) ws[wv[i]]++;
        for (i = 1; i < m; ++i) ws[i] += ws[i -1];
        for (i = n - 1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];

        t = x, x = y, y = t;
        for (x[sa[0]] = 0, p = i = 1; i < n; ++i)
            x[sa[i]] = Cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}

void change() {
    for (int i = 0; i < len * 2; ++i)
        S2[i] = S1[len * 2 - i - 1];
    S2[len * 2] = 'z' + 1;
    S2[len * 2 + 1] = 0;
}

int main() {
    int T;
    scanf("%d", &T);

    for (int xx = 0; xx < T; ++xx) {
        int x1, x2;
        scanf("%d%s", &len, S1);
        strcpy(TT, S1);
        strcat(S1, TT);
        DA(S1, len * 2 + 1, 300);
        for (int i = len * 2; i >= 0; --i) {
            if (sa[i] < len) {
                x1 = sa[i];
                break;
            }
        }

        change();
        DA(S2, len * 2 + 2, 300);
        for (int i = len * 2; i >= 0; --i) {
            if (sa[i] < len) {
                x2 = sa[i];
                break;
            }
        }

        int flag = 0;
		int k;
        for (k = 0; k < len; ++k) {
            if (S1[x1 + k] > S2[x2 + k]) 
				break;
            else if (S1[x1 + k] < S2[x2 + k]) {
                flag = 1;
                break;
            }
        }
		x2 = len - x2;
        ++x1;


        if (k < len) {
            if (flag == 0)
                printf("%d %d\n", x1, 0);
            else
                printf("%d %d\n", x2, 1);
        }
        else {
            if (x1 <= x2)
                printf("%d %d\n", x1, 0);
            else
                printf("%d %d\n", x2, 1);
        }

    }
    return 0;
}