好像还没有写过最小表示法以前都是用SAM写的
用两个指针(i)和(j)指向开头两个不同的位置,然后找到它们往后第一个不同的地方(i + k),比较大小
较大的移动指针(k + 1)位
直至到达末尾或者(k = len)
由于两指针移动次数和比较次数时同阶的,所以复杂度是(O(n))
BZOJ2882
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 300005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
int A[maxn],n,pos;
void work(){
int i = 0,j = 1,k = 0,t;
while (i < n && j < n && k < n){
t = A[(i + k) % n] - A[(j + k) % n];
if (!t) k++;
else {
if (t > 0) i += k + 1;
else j += k + 1;
if (i == j) j++;
k = 0;
}
}
pos = i < j ? i : j;
}
int main(){
n = read();
for (int i = 0; i < n; i++) A[i] = read();
work();
for (int i = 0; i < n; i++) printf("%d ",A[(pos + i) % n]);
return 0;
}