題目傳送門

做法:

• 我們知道SAM中 l[st] 陣列的意義是狀態st的最長子串的長度
• 從前一狀態到當前狀態的轉移所增加的子串數量:
• l[st] - l[slink[st]]

AC程式碼：

#include<bits/stdc++.h>
#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
 

#define abs(x)      ((x)<0 ? -(x) : x)
#define all(x)      x.begin(),x.end(
#define mk(x,y)     make_pair(x,y)
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan
 
(1.0);
const int maxm = 1e8+5;
const int maxn = 1e6+5;
const int INF = 0x3f3f3f3f;
char s[maxn];
struct SuffixAutoMation
{
	int last,cnt;
	int trans[maxn<<1][26],slink[maxn<<1],l[maxn<<1];
	int add(int x){
		int p = last,np = ++cnt;last = np;l[np] = l[p]+1;
		for(;p&&!trans[p]
 
[x];p =slink[p]) trans[p][x] = np;
		if(!p) slink[np] = 1;
		else{
			int q = trans[p][x];
			if(l[p]+1 == l[q]) slink[np] = q;
			else{
				int nq = ++cnt;l[nq] = l[p]+1;
				memcpy(trans[nq],trans[q],sizeof(trans[q]));
				slink[nq] = slink[q];
				slink[q] = slink[np] = nq;
				for(;trans[p][x] == q;p = slink[p]) trans[p][x] = nq;
			}
		}
		return l[np] - l[slink[np]];
	}
	void build(){
		ll ans = 0;
		scanf("%s",s+1);
		int len = strlen(s+1);
		last = cnt = 1;
		for(int i=1;i<=len;i++)
			ans+=add(s[i]-'a');
		printf("%lld\n",ans);
	}
	
}sam;
int main()
{
	// fin;
	IO;
	sam.build();
	return 0;
}