BZOJ3238 SA//SAM
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\n", x) #define Prl(x) printf("%lld\n",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\n", x) #define Prl(x) printf("%lld\n",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; }
