HDU 5442 Favorite Donut（后缀数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5442

题意：给出一个字符串环，要求以某点作为起点选择顺逆方向吃完整个环，使得字典序最大，如果有多种选择，先选择位置最靠前的再选取顺时针方向

思路：先按正方向将串扩成两倍，再按反方向将串扩成两倍，中间用较小的特殊符连接，然后找到后缀最大的字符串，然后利用高度数组逆着判断当前字符串是否仍满足字典序最大，注意判断该后缀所对应的位置是否满足要求

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#include <string>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long ll;const int MAXN = 401000;int sa[MAXN], height[MAXN], ran[MAXN];int r[MAXN];int t1[MAXN], t2[MAXN], c[MAXN];bool cmp(int *r, int a, int b, int l){return r[a] == r[b] && r[a + l] == r[b + l];}void da(int str[], int sa[], int ran[], int height[], int n, int m){n++;int i, j, p, *x = t1, *y = t2;for (i = 0; i < m; i++)c[i] = 0;for (i = 0; i < n; i++)c[x[i] = str[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 (j = 1; j <= n; j <<= 1){p = 0;for (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 < m; i++)c[i] = 0;for (i = 0; i < n; i++)c[x[y[i]]]++;for (i = 1; 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]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;if (p >= n)break;m = p;}int k = 0;n--;for (i = 0; i <= n; i++)ran[sa[i]] = i;for (i = 0; i < n; i++){if (k)k--;j = sa[ran[i] - 1];while (str[i + k] == str[j + k])k++;height[ran[i]] = k;}}char s[MAXN];struct node{int p, c;bool operator < (const node &rhs) const{if (p == rhs.p) return c < rhs.c;return p < rhs.p;}} ans[MAXN];int main(){int t;scanf("%d", &t);while (t--){int n;scanf("%d%s", &n, s);int len = n * 4 + 1;for (int i = 0; i < n; i++)r[i] = r[i + n] = s[i] - 'a' + 2;r[n * 2] = 1;for (int i = 1; i <= n; i++)r[n * 2 + i] = r[n * 3 + i] = s[n - i] - 'a' + 2;r[len] = 0;da(r, sa, ran, height, len, 30);// for (int i = 0; i <= len; i++)// printf("%d ", sa[i]);int num = 0;ans[num++].p = sa[len];for (int i = len - 1; i > 1; i--){//printf("************%d %d\n", sa[i], sa[i - 1]);if (height[i + 1] < n) break;if (sa[i] >= n && sa[i] <= n * 2) continue;if (sa[i] >= n * 3 + 1) continue;ans[num++].p = sa[i];}for (int i = 0; i < num; i++){if (ans[i].p <= n * 2){ans[i].p++;ans[i].c = 0;}else{ans[i].p = n - (ans[i].p - n * 2 - 1);ans[i].c = 1;}}sort(ans, ans + num);cout << ans[0].p << " " << ans[0].c << endl;}return 0;}