標籤:style os io for amp 演算法 size ad
思路:其實很簡單,就是兩個字串串連起來,中間用個特殊字元隔開,然後用尾碼數組求最長公用首碼,然後不同在兩個串中,並且最長的就是最長公用子串了。
注意的是:用第一個字串來判斷是不是在同一個字元中,剛開始用了第二個字元的長度來判斷WA了2發才發現。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7fffffff#define maxn 200010using namespace std;typedef long long ll;typedef unsigned long long ull;void radix(int *str,int *a,int *b,int n,int m){ static int count[maxn]; mem(count,0); for(int i=0; i<n; i++) ++count[str[a[i]]]; for(int i=1; i<=m; i++) count[i]+=count[i-1]; for(int i=n-1; i>=0; i--) b[--count[str[a[i]]]]=a[i];}void suffix(int *str,int *sa,int n,int m) //倍增演算法計算出尾碼數組sa{ static int rank[maxn],a[maxn],b[maxn]; for(int i=0; i<n; i++) rank[i]=i; radix(str,rank,sa,n,m); rank[sa[0]]=0; for(int i=1; i<n; i++) rank[sa[i]]=rank[sa[i-1]]+(str[sa[i]]!=str[sa[i-1]]); for(int i=0; 1<<i<n; i++) { for(int j=0; j<n; j++) { a[j]=rank[j]+1; b[j]=j+(1<<i)>=n?0:rank[j+(1<<i)]+1; sa[j]=j; } radix(b,sa,rank,n,n); radix(a,rank,sa,n,n); rank[sa[0]]=0; for(int j=1; j<n; j++) rank[sa[j]]=rank[sa[j-1]]+(a[sa[j-1]]!=a[sa[j]]||b[sa[j-1]]!=b[sa[j]]); }}void calcHeight(int *str,int *sa,int *h,int *rank,int n) //求出最長公用首碼數組h{ int k=0; h[0]=0; for(int i=0; i<n; i++) rank[sa[i]]=i; for(int i=0; i<n; i++) { k=k==0?0:k-1; if(rank[i]) while(str[i+k]==str[sa[rank[i]-1]+k]) k++; else k=0; h[rank[i]]=k; }}int a[maxn],sa[maxn],height[maxn],rank[maxn];string s,ss;int main(){ //freopen("1.txt","r",stdin); while(cin>>s>>ss) { ss=s+"#"+ss; copy(ss.begin(),ss.end(),a); int n=ss.size(),len=0; suffix(a,sa,n,256); calcHeight(a,sa,height,rank,n); for(int i=1; i<n; i++) if(height[i]>len&&((sa[i]<s.size())!=sa[i-1]<s.size())) len=height[i]; cout<<len<<endl; } return 0;}/*jworerrrrrrrreeeeeeeeeabcdstedste*/