Secretary
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1655 | Accepted: 671 |
Description
The basic condition of success of a political party, it is the good Election Programme. PSOS know about it, so they entrust the top secretary Juliet with this task. Because she wanted to make her work easier, she used her charm to talk round her friend Romeo to help her. Romeo is assistent of another political party and he was writing the programme some time ago. While writing the Programme for Juliet, he used some parts of his previous programme. When he gave the finished Programme to Juliet, they recognized that both programmes are too similar and that someone could notice it. They need to determine the longest part of text which is common to both programmes.
Input
At the first line there is a positive integer N stating the number of assignments to follow. Each assignment consists of exactly two lines of text, each of them contains at most 10000 characters. The end-of-line character is not considered to be a part of the text.
Output
Print a single line of text for each assignment. The line should contain the sentence "Nejdelsi spolecny retezec ma delku X." (The longest common part of text has X characters). Replace X with the length of the longest common substring of both texts.
Sample Input
2 Tady nejsou zadni mimozemstani. Lide tady take nejsou. Ja do lesa nepojedu. V sobotu pojedeme na vylet.
Sample Output
Nejdelsi spolecny retezec ma delku 7. Nejdelsi spolecny retezec ma delku 5.
Source
题意：求LCS长度。
代码：
1 //#include"bits/stdc++.h" 2 #include"cstdio" 3 #include"map" 4 #include"set" 5 #include"cmath" 6 #include"queue" 7 #include"vector" 8 #include"string" 9 #include"cstring" 10 #include"ctime" 11 #include"iostream" 12 #include"cstdlib" 13 #include"algorithm" 14 #define db double 15 #define ll long long 16 #define ull unsigned long long 17 #define vec vector<ll> 18 #define mt vector<vec> 19 #define ci(x) scanf("%d",&x) 20 #define cd(x) scanf("%lf",&x) 21 #define cl(x) scanf("%lld",&x) 22 #define pi(x) printf("%d\n",x) 23 #define pd(x) printf("%f\n",x) 24 #define pl(x) printf("%lld\n",x) 25 //#define rep(i, x, y) for(int i=x;i<=y;i++) 26 #define rep(i, n) for(int i=0;i<n;i++) 27 using namespace std; 28 const int N = 1e6 + 5; 29 const int mod = 1e9 + 7; 30 const int MOD = mod - 1; 31 const int inf = 0x3f3f3f3f; 32 const db PI = acos(-1.0); 33 const db eps = 1e-10; 34 int sa[N]; 35 int rk[N]; 36 int tmp[N]; 37 int lcp[N]; 38 int n,k,T; 39 bool cmp(int i,int j){ 40 if(rk[i] != rk[j]) return rk[i]<rk[j]; 41 else 42 { 43 int ri=i+k<=n?rk[i+k]:-1; 44 int rj=j+k<=n?rk[j+k]:-1; 45 return ri<rj; 46 } 47 } 48 void bulid(string s,int *sa) 49 { 50 n=(int)s.size(); 51 for(int i=0;i<=n;i++){ 52 sa[i]=i; 53 rk[i]=i<n?s[i]:-1; 54 } 55 for(k=1;k<=n;k*=2){ 56 sort(sa,sa+n+1,cmp); 57 tmp[sa[0]]=0; 58 for(int i=1;i<=n;i++){ 59 tmp[sa[i]]=tmp[sa[i-1]]+(cmp(sa[i-1],sa[i])?1:0); 60 } 61 for(int i=0;i<=n;i++){ 62 rk[i]=tmp[i]; 63 } 64 } 65 } 66 void LCP(string s,int *sa,int *lcp){ 67 n=(int)s.size(); 68 for(int i=0;i<=n;i++) rk[sa[i]]=i; 69 int h=0; 70 lcp[0]=0; 71 for(int i=0;i<n;i++){ 72 int j=sa[rk[i]-1]; 73 for (h ? h-- : 0; j + h < n&&i + h < n&&s[j + h] == s[i + h]; h++); 74 lcp[rk[i]-1] = h; 75 } 76 } 77 int main() 78 { 79 ios::sync_with_stdio(0); 80 cin>>T; 81 getchar(); 82 while(T--) 83 { 84 string s,t; 85 getline(cin,s);//read a line 86 getline(cin,t); 87 int s1=(int)s.size(); 88 s+='$'+t; 89 n=(int)s.size(); 90 bulid(s,sa); 91 LCP(s,sa,lcp); 92 int ma=0; 93 for(int i = 0;i < n; i++){ 94 if((sa[i]<s1)!=(sa[i+1]<s1)) ma=max(ma,lcp[i]); 95 } 96 printf("Nejdelsi spolecny retezec ma delku %d.\n",ma); 97 } 98 return 0; 99 }