POJ 3693 Maximum repetition substring（後綴數組神題)

POJ 3693 Maximum repetition substring

題目鏈接

題意：給定一個字符串，求出其子串中，重復次數最多的串，如果有相同的，輸出字典序最小的

思路：枚舉長度l，把字符串按l分段，這樣對於長度為l的字符串，肯定會包含一個分段位置，這樣一來就可以在每個分段位置，往後做一次lcp，求出最大匹配長度，然後如果匹配長度有剩余，看剩余多少，就往前多少位置再做一次lcp，如果匹配出來長度更長，匹配次數就加1，這樣就可以枚舉過程中保存下答案了

這樣問題還有字典序的問題，這個完全可以利用sa數組的特性，從字典序最小往大枚舉，直到出現一個符合的位置就輸出結束

代碼：

#include 
#include 
#include 
using namespace std;

typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXLEN = 200005;

struct Suffix {

    int s[MAXLEN];
    int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;
    int rank[MAXLEN], height[MAXLEN];
    int best[MAXLEN][20];

    int len;
    char str[MAXLEN];
    int ans[MAXLEN], an;

    void build_sa(int m) {
	n++;
	int i, *x = t, *y = t2;
	for (i = 0; i < m; i++) c[i] = 0;
	for (i = 0; i < n; i++) c[x[i] = s[i]]++;
	for (i = 1; i < m; i++) c[i] += c[i - 1];
	for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
	for (int k = 1; k <= n; k <<= 1) {
	    int p = 0;
	    for (i = n - k; i < n; i++) y[p++] = i;
	    for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;
	    for (i = 0; i < m; i++) c[i] = 0;
	    for (i = 0; i < n; i++) c[x[y[i]]]++;
	    for (i = 0; i < m; i++) c[i] += c[i - 1];
	    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
	    swap(x, y);
	    p = 1; x[sa[0]] = 0;
	    for (i = 1; i < n; i++)
		x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p++;
	    if (p >= n) break;
	    m = p;
	}
	n--;
    }

    void getHeight() {
	int i, j, k = 0;
	for (i = 1; i <= n; i++) rank[sa[i]] = i;
	for (i = 0; i < n; i++) {
	    if (k) k--;
	    int j = sa[rank[i] - 1];
	    while (s[i + k] == s[j + k]) k++;
	    height[rank[i]] = k;
	}
    }

    void initRMQ() {
	for (int i = 0; i < n; i++) best[i][0] = height[i + 1];
	for (int j = 1; (1< R) swap(L, R);
	L++;
	int k = 0;
	while ((1<<(k + 1)) <= R - L + 1) k++;
	return min(best[L][k], best[R - (1<= 0 && tmp % l && lcp(v, v + l) >= tmp)
		    ti++;
		if (ti > Max) {
		    an = 0;
		    ans[an++] = l;
		    Max = ti;
		}
		else if (ti == Max)
		    ans[an++] = l;
	    }
	}
	int ans_v, ans_l;
	for (int i = 1; i <= n; i++) {
	    int flag = 0;
	    for (int j = 0; j < an; j++) {
		int tmp = ans[j];
		if (lcp(sa[i], sa[i] + tmp) >= (Max - 1) * tmp) {
		    ans_v = sa[i];
		    ans_l = Max * tmp;
		    flag = 1;
		}
	    }
	    if (flag) break;
	}
	for (int i = 0; i < ans_l; i++)
	    printf("%c", str[ans_v + i]);
	printf("\n");
    }

} gao;

int main() {
    int cas = 0;
    while(~scanf("%s", gao.str) && gao.str[0] != '#') {
	printf("Case %d: ", ++cas);
	gao.solve();
    }
    return 0;
}