spoj694/705 Distinct Substrings - 后缀数组

来源：互联网发布：编程的8大算法编辑：程序博客网时间：2024/06/05 16:14

题目链接：http://acm.hust.edu.cn/vjudge/problem/19282

题目大意：求不同子串的个数

解题思路：后缀数组..
suffix(i)对子串个数所做的贡献为len-sa[i]+1，因为要求要不同的，所以减去与它串重复的子串个数h[i]。即每个后缀对答案的贡献为len-sa[i]+1-h[i];(***)

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;#define maxn 1100int sa[maxn],rk[maxn],y[maxn],len;int Rsort[maxn],wr[maxn],h[maxn];char s[maxn];bool cmp(int x,int y,int ln){return (wr[x]==wr[y])&&(wr[x+ln]==wr[y+ln]);}void get_h(){int i,k=0;for (i=1;i<=len;i++){int j=sa[rk[i]-1];if (k>0) k--;while (s[i+k]==s[j+k]) k++;h[rk[i]]=k;}}void get_sa(){int i,k,p,ln,m=130;for (i=1;i<=len;i++) rk[i]=s[i];for (i=0;i<=m;i++) Rsort[i]=0;for (i=1;i<=len;i++) Rsort[rk[i]]++;for (i=1;i<=m;i++) Rsort[i]+=Rsort[i-1];for (i=len;i>=1;i--) sa[Rsort[rk[i]]--]=i;p=0;ln=1;while (p<len){for (k=0,i=len-ln+1;i<=len;i++) y[++k]=i;for (i=1;i<=len;i++) if (sa[i]-ln>0) y[++k]=sa[i]-ln;for (i=1;i<=len;i++) wr[i]=rk[y[i]];for (i=0;i<=m;i++) Rsort[i]=0;for (i=1;i<=len;i++) Rsort[wr[i]]++;for (i=1;i<=m;i++) Rsort[i]+=Rsort[i-1];for (i=len;i>=1;i--) sa[Rsort[wr[i]]--]=y[i];memcpy(wr,rk,sizeof(wr));p=1;rk[sa[1]]=1;for (i=2;i<=len;i++){if (!cmp(sa[i],sa[i-1],ln)) p++;rk[sa[i]]=p;}m=p;ln*=2;}sa[0]=s[0]=0;}int main(){int T,i;scanf("%d\n",&T);while (T--){scanf("%s",s+1);len=strlen(s+1);get_sa();get_h();LL ans=0;h[1]=0;for (i=1;i<=len;i++){LL ls=sa[i];ans+=len-ls+1-h[i];}printf("%lld\n",ans);}return 0;}

0 0