Description
Input
Output
Sample Input
input | output |
---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
Source
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <bitset> #include <algorithm> #include <climits> using namespace std; #define LS 2*i #define RS 2*i+1 #define UP(i,x,y) for(i=x;i<=y;i++) #define DOWN(i,x,y) for(i=x;i>=y;i--) #define MEM(a,x) memset(a,x,sizeof(a)) #define W(a) while(a) #define gcd(a,b) __gcd(a,b) #define LL long long #define N 2222 #define MOD 1000000007 #define INF 0x3f3f3f3f #define EXP 1e-8 int wa[N],wb[N],wsf[N],wv[N],sa[N]; int rank[N],height[N],s[N],a[N]; char str[N],str1[N],str2[N]; //sa:字典序中排第i位的起始位置在str中第sa[i] //rank:就是str第i个位置的后缀是在字典序排第几 //height:字典序排i和i-1的后缀的最长公共前缀 int cmp(int *r,int a,int b,int k) { return r[a]==r[b]&&r[a+k]==r[b+k]; } void getsa(int *r,int *sa,int n,int m)//n要包含末尾添加的0 { int i,j,p,*x=wa,*y=wb,*t; for(i=0; i<m; i++) wsf[i]=0; for(i=0; i<n; i++) wsf[x[i]=r[i]]++; for(i=1; i<m; i++) wsf[i]+=wsf[i-1]; for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i; p=1; j=1; for(; p<n; j*=2,m=p) { for(p=0,i=n-j; i<n; i++) y[p++]=i; for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j; for(i=0; i<n; i++) wv[i]=x[y[i]]; for(i=0; i<m; i++) wsf[i]=0; for(i=0; i<n; i++) wsf[wv[i]]++; for(i=1; i<m; i++) wsf[i]+=wsf[i-1]; for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i]; t=x; x=y; y=t; x[sa[0]]=0; for(p=1,i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++; } } void getheight(int *r,int n)//n不保存最后的0 { int i,j,k=0; for(i=1; i<=n; i++) rank[sa[i]]=i; for(i=0; i<n; i++) { if(k) k--; else k=0; j=sa[rank[i]-1]; while(r[i+k]==r[j+k]) k++; height[rank[i]]=k; } } int main() { int len,n=0,i,j,k; W(~scanf("%s",str)) { n = 0; len = strlen(str); UP(i,0,len-1) s[n++] = str[i]; s[n++] = 200; DOWN(i,len-1,0) s[n++] = str[i]; s[n] = 0; getsa(s,sa,n+1,300); getheight(s,n); int ans = 1,x=0; UP(i,1,n-1) { int minn = min(sa[i],sa[i-1]); int maxn = max(sa[i],sa[i-1]); if(minn>=len || maxn<len) continue;//要分别在两个串中 if(minn+height[i]!=n-maxn) continue;//minn+height[i]是两者公共前缀的最后一个,而这个对应的位置是在后一串的n-(n-sa[k]),也就是n-maxn位置 if(height[i]>ans) { ans = height[i]; x = minn; } else if(height[i]==ans) { x = min(minn,x); } } for(i=x; ans--; i++) printf("%c",str[i]); puts(""); } }
原文地址:http://blog.csdn.net/libin56842/article/details/46281255