Substrings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10244 | Accepted: 3515 |
Description
You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.
Output
There should be one line per test case containing the length of the largest string found.
Sample Input
2 3 ABCD BCDFF BRCD 2 rose orchid
Sample Output
2 2
Source
#include<cstdio> #include<cstring> using namespace std; const int maxn=20200; int wa[maxn],wb[maxn],wv[maxn], ws[maxn]; int cmp(int *r,int a,int b,int l){ return r[a]==r[b] && r[a+l]==r[b+l]; } void da(int *r,int *sa,int n,int m){ int i,j,p,*x=wa,*y=wb,*t; for(i=0;i<m;i++) ws[i]=0; for(i=0;i<n;i++) ws[x[i]=r[i]]++; for(i=1;i<m;i++) ws[i]+=ws[i-1]; for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i; for(j=1,p=1;p<n;j<<=1,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++) ws[i]=0; for(i=0;i<n;i++) ws[wv[i]]++; for(i=1;i<m;i++) ws[i]+=ws[i-1]; for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i]; for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; } } int rank[maxn],height[maxn]; void calheight(int *r,int *sa,int n){ int i,j,k=0; for(i=1;i<=n;i++) rank[sa[i]]=i; for(i=0;i<n;height[rank[i++]]=k) for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++) ; } int r[maxn],sa[maxn]; int who[maxn],yes[110]={0},cnt=0; int len,n; int check(int mid){ int i,j,k,t,s; for(i=2;i<=len;i=j+1){ for(;height[i]<mid && i<=len;i++) ; for(j=i;height[j]>=mid;j++) ; if(j-i+1<n) continue; cnt++; s=0; for(k=i-1;k<j;k++) if((t=who[sa[k]])!=0 && yes[t]!=cnt) yes[t]=cnt,s++; if(s>=n) return 1; } return 0; } char st[110]; int main() { //freopen("input.txt","r",stdin); int i,j,k,min,mid,max,nn; scanf("%d",&nn); while(nn-->0) { scanf("%d",&n); len=0; for(i=1;i<=n;i++) { scanf("%s",st); k=strlen(st); for(j=0;j<k;j++) { r[j+len]=st[j]+200; who[j+len]=i; } r[len+k]=2*i-1; who[len+k]=0; len+=k+1; for(j=0;j<k;j++) { r[j+len]=st[k-1-j]+200; who[j+len]=i; } r[len+k]=2*i; who[len+k]=0; len+=k+1; } len--; r[len]=0; da(r,sa,len+1,328); calheight(r,sa,len); height[len+1]=-1; min=1;max=100; while(min<=max) { mid=(min+max)>>1; if(check(mid)) min=mid+1; else max=mid-1; } if(n==1) max=len/2; printf("%d\n",max); } return 0; }
2,标准库函数strncpy()，可以将一字符串的一部分拷贝到另一个字符串中。strncpy()函数有3个参数：第一个参数是目录字符串；第二个参数是源字符串；第三个参数是一个整数，代表要从源字符串拷贝到目标字符串中的字符数
题意：给定n个串，求一个最大子串长度，使得它或者它的逆向串在每个串中出现。
思路：先求出最短的串sho[]，然后枚举答案长度ans（len~1）。于是sho[]就被分成了len-ans+1个子串pos[],再分别求出这些字串的反串inv[],判断pos[]和inv[]是否都出现在所有的串中。
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int n; char str[110][110],sho[110]; int main(){ //freopen("input.txt","r",stdin); int t; scanf("%d",&t); while(t--){ scanf("%d",&n); int len=110,k; for(int i=0;i<n;i++){ scanf("%s",str[i]); int tmp=strlen(str[i]); if(tmp<len){ len=tmp; k=i; // 找出最短的串。 } } strcpy(sho,str[k]); int ans=len,flag=0; char pos[110],inv[110]; while(ans>0){ for(int i=0;i<=len-ans;i++){ strncpy(pos,sho+i,ans); // 求出正序子串pos[]。 for(int j=0;j<ans;j++) inv[j]=pos[ans-1-j]; // 求出反序子串inv[]。 pos[ans]=inv[ans]='\0'; flag=1; for(int j=0;j<n;j++) if(!strstr(str[j],pos) && !strstr(str[j],inv)){ // 判断pos[]，inv[]是否都出现在n个串中。运用strstr函数。 flag=0; break; } if(flag) break; } if(flag) break; else ans--; } printf("%d\n",ans); } return 0; }
3,KMP
/* |
#include <cstdio> #include <cstring> using namespace std; const int maxN = 100 + 5; const int maxL = 100 + 5; //minLen, minNum : 最短串 int caseNum, n, minLen, minNum; int next[maxL]; //串 反串 char str[maxN][maxL], restr[maxN][maxL]; void reverseString(int i, int len){ for(int j=0; j<len; j++) restr[i][j] = str[i][len - j -1]; restr[i][len] = 0; } //以start为起点的模式 void preKMP(int start){ int j, k; next[start] = start - 1; j = start, k = start - 1; while(j < minLen - 1){ while(k > start - 1 && str[minNum][j] != str[minNum][k]){ k = next[k]; } j++, k++; if(str[minNum][j] == str[minNum][k]) next[j] = next[k]; else next[j] = k; } } int KMP(int t, int start){ int i, j, ansMax, tmpMax = start; int tLen = strlen(str[t]); i = 0; j = start; while(i < tLen && j < minLen){ while(j >= start && str[t][i] != str[minNum][j]){ j = next[j]; } i++, j++; //记录最大的j if(tmpMax < j) tmpMax = j; } ansMax = tmpMax - start; tmpMax = start; i = 0; j = start; while(i < tLen && j <minLen){ while(j >= start && restr[t][i] != str[minNum][j]){ j = next[j]; } i++, j++; if(tmpMax < j) tmpMax = j; } //串与反串的最大j if(ansMax < tmpMax - start) ansMax = tmpMax - start; return ansMax; } int main(){ scanf("%d", &caseNum); while(caseNum--){ scanf("%d", &n); //注意n == 1 ， 为此WA了两次！！ if(n == 1){ scanf("%s", str[0]); printf("%d\n", strlen(str[0])); continue; } minLen = maxL; for(int i=0; i<n; i++){ scanf("%s", str[i]); int len = strlen(str[i]); reverseString(i, len); if(len < minLen){ minLen = len; minNum = i; } } int tmpMin = maxL, tmp, tmpMax = -1; //枚举起点 for(int i=0; i<minLen; i++){ preKMP(i); tmpMin = maxL; for(int j=0; j<n; j++){ if(j == minNum) continue; tmp = KMP(j, i); if(tmp < tmpMin) tmpMin = tmp; } if(tmpMax < tmpMin) tmpMax = tmpMin; } printf("%d\n", tmpMax); } return 0; }