BZOJ3238 SA//SAM

题目链接

题意：求两两后缀的LCP的和

做法一：

很容易想到后缀数组，Height数组表示的是排名相邻两后缀的LCP

但是可以意识到任意两后缀的LCP是他们之间Height的最小值，所以只要计算出每个点的Height作为最小值覆盖到最左和最右的点，那么r[i] - i + 1--- i - l[i] + 1这个数量的区间形成的后缀都是以当前Height为最小值的

这是一个经典的单调栈//DP问题，两种方法都可以线性求出来，需要特别注意的是两种方法都需要注意相同Height回来带来重复计算，例如1,1,1的height，直接计算的话每个值都管到[1,3]，因此要作一个左闭右开或左开右闭的情况，即[1,3],[2,3],[3,3]

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d
", x)
#define Prl(x) printf("%lld
",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
char str[maxn];
int sa[maxn],rak[maxn],tex[maxn],tp[maxn],Height[maxn];
void GetHeight(){
    int j,k = 0;
    for(int i = 1; i <= N ; i ++){
        if(k) k--;
        int j = sa[rak[i] - 1];
        while(str[i + k] == str[j + k]) k++;
        Height[rak[i]] = k;
    }
} 
void Qsort(){
    for(int i = 0; i <= M; i ++) tex[i] = 0;
    for(int i = 1; i <= N; i ++) tex[rak[i]]++;
    for(int i = 1; i <= M; i ++) tex[i] += tex[i - 1];
    for(int i = N; i >= 1; i --) sa[tex[rak[tp[i]]]--] = tp[i];
}
void SA(){
    for(int i = 1; i <= N ; i ++) rak[i] = str[i] - 'a' + 1,tp[i] = i;
    Qsort();
    for(int w = 1,p = 0; p < N; w <<= 1,M = p){
        p = 0;
        for(int i = 1; i <= w; i ++) tp[++p] = N - w + i;
        for(int i = 1; i <= N; i ++) if(sa[i] > w) tp[++p] = sa[i] - w;
        Qsort();
        swap(tp,rak);
        rak[sa[1]] = p = 1;
        for(int i = 2; i <= N; i ++){
            rak[sa[i]] = (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])?p:++p;
        }
    }
}
int Stack[maxn],top;
int l[maxn],r[maxn];
int main(){
    scanf("%s",str + 1); N = strlen(str + 1);
    M = 122;
    SA(); GetHeight();
    for(int i = 1; i <= N ; i ++){
        while(top && Height[Stack[top]] > Height[i]) top--;
        l[i] = Stack[top] + 1;
        Stack[++top] = i;
    }
    top = 0; Stack[0] = N + 1;
    for(int i = N; i >= 1; i --){
        while(top && Height[Stack[top]] >= Height[i]) top--;
        r[i] = Stack[top] - 1;
        Stack[++top] = i;
    }
    LL sum = 1ll * (N - 1) * (1 + N) * N / 2;
    for(int i = 1; i <= N; i ++){
        //cout << l[i] << " " << r[i] << endl;
        sum -= 1ll * 2 * (r[i] - i + 1) * (i - l[i] + 1) * Height[i];
    }
    Prl(sum);
    return 0;
}

做法二：

SAM一个重要的性质就是parent树上，两节点的最长公共后缀是他们LCA表示的最长字符串

所以我们将字符串逆转，就变成了求N个代表前缀的关键点两两之间的len[LCA]的和，事实上right数组表示的就是子树下有多少关键点，

所以可以每个点的关键点对数就是num[u] * (num[u] - 1) - Σnum[v] - (num[v] - 1) (v 为 u的儿子)

对数可以O(n)求出，最后统计的时候减去对数 * len[i]即可。

也可以用num[i] * (num[i] - 1) * (len[i] - len[fa[i]) 作为减去的贡献，理解起来稍微抽象一点，就是计入所有贡献之后减去LCA为父亲及以上的点的贡献

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d
", x)
#define Prl(x) printf("%lld
",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 5e5 + 10;
const int maxc = 26;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int len[maxn << 1],fa[maxn << 1],son[maxn << 1][maxc];
LL num[maxn << 1];
int size,last;
LL sum;
void Init(){
    size = last = 1;
}
void insert(char c){
    int s = c - 'a';
    int p = last,np = ++size;last = np; 
    //cout << np << endl;
    len[np] = len[p] + 1; num[np] = 1; 
    for(;p && !son[p][s]; p = fa[p]) son[p][s] = np;
    if(!p) fa[np] = 1;
    else{
        int q = son[p][s];
        if(len[p] + 1 == len[q]) fa[np] = q;
        else{
            int nq = ++size; len[nq] = len[p] + 1;
            memcpy(son[nq],son[q],sizeof(son[q]));
            fa[nq] = fa[q]; fa[q] = fa[np] = nq; 
            for(;son[p][s] == q && p;p = fa[p]) son[p][s] = nq;
        }
    }
}
int A[maxn << 1],tmp[maxn << 1];
void Qsort(){
    for(int i = 1; i <= size; i ++) tmp[len[i]]++;
    for(int i = 1; i <= size; i ++) tmp[i] += tmp[i - 1];
    for(int i = 1; i <= size; i ++) A[tmp[len[i]]--] = i;
}
char str[maxn];
LL dp[maxn << 1];
int main(){
    scanf("%s",str); N = strlen(str);
    reverse(str,str + N);
    sum = 1ll * (N - 1) * (N + 1) * N / 2; Init();
    for(int i = 0 ; i < N ; i ++) insert(str[i]);
    Qsort();
    for(int i = size; i >= 1; i --) num[fa[A[i]]] += num[A[i]];
    for(int i = 2; i <= size; i ++){
        sum -= (len[i] - len[fa[i]]) * num[i] * (num[i] - 1);
    }
//    for(int i = 2; i <= size; i ++) sum -= len[i] * dp[i];
    Prl(sum);
    return 0;
}