标签：scan scanf nta add max ota case using bsp
Given a string, we need to find the total number of its distinct substrings.
T- number of test cases. T<=20;
Each test case consists of one string, whose length is <= 1000
For each test case output one number saying the number of distinct substrings.
Sample Input:
2
CCCCC
ABABA
Sample Output:
5
9
Explanation for the testcase with string ABABA:
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA
Thus, total number of distinct substrings is 9.
求出大写的字符串里不同的子串。默写了一遍后缀自动机。今天主要是练习后缀数组。
#include<cstdio> #include<cstdlib> #include<iostream> #include<cstring> #include<algorithm> using namespace std; const int maxn=10000; struct SAM { int ch[maxn][26],fa[maxn],maxlen[maxn],Last,sz; void init() { sz=Last=1; fa[1]=maxlen[1]=0; memset(ch[1],0,sizeof(ch[1])); } void add(int x) { int np=++sz,p=Last;Last=np; memset(ch[np],0,sizeof(ch[np])); maxlen[np]=maxlen[p]+1; while(p&&!ch[p][x]) ch[p][x]=np,p=fa[p]; if(!p) fa[np]=1; else { int q=ch[p][x]; if(maxlen[p]+1==maxlen[q]) fa[np]=q; else { int nq=++sz; memcpy(ch[nq],ch[q],sizeof(ch[q])); maxlen[nq]=maxlen[p]+1; fa[nq]=fa[q]; fa[q]=fa[np]=nq; while(p&&ch[p][x]==q) ch[p][x]=nq,p=fa[p]; } } } }; SAM Sam; int main() { char chr[maxn]; int T,ans,i,L; scanf("%d",&T); while(T--){ Sam.init();ans=0; scanf("%s",chr); L=strlen(chr); for(i=0;i<L;i++) Sam.add(chr[i]-‘A‘); for(i=1;i<=Sam.sz;i++) ans+=Sam.maxlen[i]-Sam.maxlen[Sam.fa[i]]; printf("%d\n",ans); } return 0; }
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int maxn=10000; char ch[maxn]; int L; struct SA { int cntA[maxn],cntB[maxn],A[maxn],B[maxn]; int rank[maxn],sa[maxn],tsa[maxn],ht[maxn];void sort() { for (int i = 0; i < 256; i ++) cntA[i] = 0; for (int i = 1; i <= L; i ++) cntA[ch[i]] ++; for (int i = 1; i < 256; i ++) cntA[i] += cntA[i - 1]; for (int i = L; i; i --) sa[cntA[ch[i]] --] = i; rank[sa[1]] = 1; for (int i = 2; i <= L; i ++){ rank[sa[i]] = rank[sa[i - 1]]; if (ch[sa[i]] != ch[sa[i - 1]]) rank[sa[i]] ++; } for (int l = 1; rank[sa[L]] < L; l <<= 1){ for (int i = 0; i <= L; i ++) cntA[i] = 0; for (int i = 0; i <= L; i ++) cntB[i] = 0; for ( int i = 1; i <= L; i ++){ cntA[A[i] = rank[i]] ++; cntB[B[i] = (i + l <= L) ? rank[i + l] : 0] ++; } for (int i = 1; i <= L; i ++) cntB[i] += cntB[i - 1]; for (int i = L; i; i --) tsa[cntB[B[i]] --] = i; for (int i = 1; i <= L; i ++) cntA[i] += cntA[i - 1]; for (int i = L; i; i --) sa[cntA[A[tsa[i]]] --] = tsa[i]; rank[sa[1]] = 1; for (int i = 2; i <= L; i ++){ rank[sa[i]] = rank[sa[i - 1]]; if (A[sa[i]] != A[sa[i - 1]] || B[sa[i]] != B[sa[i - 1]]) rank[sa[i]] ++; } } } void getheight() { for (int i = 1, j = 0; i <= L; i ++){ if (j) j --; while (ch[i + j] == ch[sa[rank[i] - 1] + j]) j ++; ht[rank[i]] = j; } } }; SA Sa; int main() { int T,ans,i; scanf("%d",&T); while(T--){ ans=0; scanf("%s",ch+1); L=strlen(ch+1); Sa.sort(); Sa.getheight(); for(i=1;i<=L;i++) ans+=L-Sa.sa[i]+1-Sa.ht[i]; printf("%d\n",ans); } return 0; }
SPOJ705 Distinct Substrings （后缀自动机&后缀数组）
标签：scan scanf nta add max ota case using bsp
原文地址：http://www.cnblogs.com/hua-dong/p/8016222.html
