poj 3294 Life Forms （后缀数组）

Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 15125   Accepted: 4437 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 3 abcdefg bcdefgh cdefghi 3 xxx yyy zzz 0 Sample Output bcdefg cdefgh ? Source Waterloo Local Contest, 2006.9.30

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题目大意：求不小于K个字符串中的最长公共子串

题解：后缀数组

将N个串连接起来，然后二分答案对height分组，看某组中是否存在K个不同串的子串。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 200003
using namespace std;
int sa[N],rank[N],height[N],xx[N],yy[N],*x,*y;
int n,m,len,p,vis[N],cnt,q[N],b[N],a[N],ans,pos[N];
char s[N];
void init()
{
	memset(sa,0,sizeof(sa));
	memset(rank,0,sizeof(rank));
	memset(height,0,sizeof(height));
	memset(b,0,sizeof(b));
	memset(q,0,sizeof(q));
	memset(vis,0,sizeof(vis));
}
int cmp(int i,int j,int l)
{
	return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);
}
void get_SA()
{
	x=xx; y=yy; m=200;
	for (int i=1;i<=len;i++) b[x[i]=a[i]]++;
	for (int i=1;i<=m;i++) b[i]+=b[i-1];
	for (int i=len;i>=1;i--) sa[b[x[i]]--]=i;
    for (int k=1;k<=len;k<<=1) {
    	p=0;
    	for (int i=len-k+1;i<=len;i++) y[++p]=i;
    	for (int i=1;i<=len;i++)
    	 if (sa[i]>k) y[++p]=sa[i]-k;
    	for (int i=1;i<=m;i++) b[i]=0;
    	for (int i=1;i<=len;i++) b[x[y[i]]]++;
    	for (int i=1;i<=m;i++) b[i]+=b[i-1];
    	for (int i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i];
    	swap(x,y); p=2; x[sa[1]]=1;
    	for (int i=2;i<=len;i++) 
    	 x[sa[i]]=cmp(sa[i],sa[i-1],k)?p-1:p++;
    	if (p>len) break;
    	m=p+1;
	}
	p=0;
	for (int i=1;i<=len;i++) rank[sa[i]]=i;
	for (int i=1;i<=len;i++) {
		if (rank[i]==1) continue;
		int j=sa[rank[i]-1];
		while (i+p<=len&&j+p<=len&&a[i+p]==a[j+p]) p++;
		height[rank[i]]=p;
		p=max(p-1,0);
	}
}
int pd(int x)
{
	memset(vis,0,sizeof(vis));
	int size=1; vis[pos[sa[1]]]=1;
	int last=0;
	for (int i=2;i<=len;i++)
	 if (height[i]>=x)
	 {
	  int t=pos[sa[i]];
	  if (!vis[t]) 
	   vis[t]++,size++;
     }
     else {
     	if (size>=n/2+1) return 1;
     	for (int j=last;j<=i-1;j++)
     	 vis[pos[sa[j]]]=0;
     	size=1; last=i; vis[pos[sa[i]]]=1;
	 }
	if (size>=n/2+1) return 1;
	return 0;
}
void solve()
{
	memset(vis,0,sizeof(vis));
	int size=1; int last=0; vis[pos[sa[1]]]=1;
	for (int i=2;i<=len;i++)
	 if (height[i]>=ans) {
	 	int t=pos[sa[i]];
	    if (!vis[t]) 
	      vis[t]++,size++;
	 }
	 else {
	 	if (size>=n/2+1) q[++cnt]=sa[i-1];
	 	for (int j=last;j<=i-1;j++)
     	 vis[pos[sa[j]]]=0;
     	size=1; last=i; vis[pos[sa[i]]]=1;
	 }
	if (size>=n/2+1) q[++cnt]=sa[n];
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("my.out","w",stdout);
	while(true) {
		init();
		scanf("%d",&n);
		if (!n) break; len=0;
		for (int i=1;i<=n;i++) {
			scanf("%s",s+1);
			m=strlen(s+1);
			if (i!=1) a[++len]=i+27;
			for (int j=1;j<=m;j++)
			 a[++len]=s[j]-'a'+1,pos[len]=i;
		}
		get_SA();
		int l=1; int r=len; ans=0;
		while (l<=r) {
			int mid=(l+r)/2;
			if (pd(mid)) ans=max(ans,mid),l=mid+1;
			else r=mid-1;
		}
		cnt=0;
	    solve(); bool pd=false;
	    if (!ans) printf("?\n");
	    else {
	    	for (int i=1;i<=cnt;i++){
		    	 if (pos[q[i]]==pos[q[i]+ans-1])
		    	 {
				  pd=true;
		    	  for (int j=q[i];j<=q[i]+ans-1;j++)
		    	    printf("%c",a[j]+'a'-1);
		    	  printf("\n");
		         }
	        }
	        if (!pd) printf("?\n");
		}
		printf("\n");
	}
}