POJ3294:Life Forms(后缀数组)
来源:互联网 发布:rmvx存档修改器优化版 编辑:程序博客网 时间:2024/05/07 12:18
Description
You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.
The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant's life forms ended up with a large fragment of common DNA.
Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.
Input
Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.
Output
For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.
Sample Input
3abcdefgbcdefghcdefghi3xxxyyyzzz0
Sample Output
bcdefgcdefgh?
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 1000005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8int wa[N],wb[N],wsf[N],wv[N],sa[N];int rank[N],height[N],s[N],a[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; }}char str[N];int len[105],size,ans[N];bool vis[105];int check(int mid,int n,int k){ int i,j; int size = 0,cnt = 0; MEM(vis,false); for(i = 1; i<=n; i++) { if(height[i]>=mid) { for(j = 1; j<=k; j++) { //把sa[i-1]或sa[i]所在的字符串给标记,同样的串不重复累加 if(sa[i]>len[j-1]&&sa[i]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true; if(sa[i-1]>len[j-1]&&sa[i-1]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true; } } else { if(cnt>k/2) ans[++size] = sa[i-1]; cnt = 0; MEM(vis,false); } } if(cnt>k/2) ans[++size] = sa[n]; if(size) { ans[0] = size; return 1; } return 0;}int main(){ int n,k,i,j,flag = 0; while(~scanf("%d",&k),k) { n = 0; size = 0; for(i = 1; i<=k; i++) { scanf("%s",str+n); for(; str[n]!='\0'; n++) s[n] = str[n]; s[n] = '#'+i; len[++size] = n; n++; } s[n-1] = 0; getsa(s,sa,n,255); getheight(s,n-1); int l=1,r=n,mid; while(l<=r) { mid = (l+r)/2; if(check(mid,n,k)) l = mid+1; else r = mid-1; } if(flag) puts(""); flag = 1; if(l==1) puts("?"); else { for(i = 1; i<=ans[0]; i++) { for(j = ans[i]; j<ans[i]+l-1; j++) printf("%c",s[j]); puts(""); } } } return 0;}
- POJ3294:Life Forms(后缀数组)
- 后缀数组 - poj3294 Life Forms
- poj3294 Life Forms - 后缀数组
- POJ3294 Life Forms 【后缀数组】
- 【POJ3294】Life Forms【后缀数组】【二分】
- poj3294 Life Forms(后缀数组)
- [POJ3294]Life Forms(后缀数组+二分)
- POJ3294-Life Forms(后缀数组)
- poj3294 Life Forms(后缀数组+二分答案)
- POJ3294——Life Forms 后缀数组
- POJ3294---Life Forms(后缀数组,二分+给后缀分组)
- Life Forms poj3294
- POJ3294--Life Forms
- poj3294 Life Forms
- poj3294 Life Forms
- POJ3294-Life Forms
- Poj3294 Life Forms
- poj3294 Life Forms(后缀数组+大于k/2个字符串中含有的最长公共子串)
- 陈怡暖:纽约联储主席杜德利讲话惊泄天机
- cocos2d-x 游戏实战经验(三) 多分辨率的自适应(上)
- 合并果子,优先队列的使用
- 归纳决策树ID3(Java实现)
- shell循环,for 、while、until
- POJ3294:Life Forms(后缀数组)
- IOS界面(图片)翻转
- 【应用篇】Activiti外置表单简单应用(三)
- VS2008 使用 occi 连接 Oracle 服务器- 不用安装客户端
- asp.net发送电子邮件
- 串口调试小节之一 串口硬件连线
- Activity中的四种启动模式
- android AsyncTask介绍
- Java日历