标签：
对应POJ 题目：点击打开链接
Description
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.
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.
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.
3 abcdefg bcdefgh cdefghi 3 xxx yyy zzz 0
bcdefg cdefgh ?
思路：就是后缀数组求多字符串的最长公共子串，height数组分组+二分答案求上界。细节上，求得一组后前缀后，要判断是否含有分隔符。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MS(x, y) memset(x, y, sizeof(x))
const int MAXN = 100000+2000;
const int INF = 1<<30;
int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];
int rank[MAXN],r[MAXN],sa[MAXN],height[MAXN];
char str[1005];
int vis[1005], ID[1005];
int block[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++;
}
return;
}
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-1; height[rank[i++]] = k)
for(k ? k-- : 0,j=sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
return;
}
int main()
{
//freopen("in.txt", "r", stdin);
int n;
scanf("%d", &n);
while(n)
{
int i, j, k;
MS(rank, 0);
MS(sa, 0);
MS(wa, 0);
MS(wb, 0);
MS(ws, 0);
MS(wv, 0);
MS(r, 0);
MS(height, 0);
MS(block, 0);
MS(ID, 0);
int len = 1, tmp_l, maxn = 0;
int left = 1, right = INF;
for(i=0; i<n; i++){//把所有字符串连成一个用分隔符分隔的字符串
scanf("%s", str);
tmp_l = strlen(str);
if(tmp_l < right) right = tmp_l;//二分答案的右边界为最短字符串的长度
int k;
for(j=len, k=0; k<tmp_l; j++, k++){
block[j] = i;//下标为j的字符所在的是第i个字符串
r[j] = str[k] - 'a' + 1;
if(r[j] > maxn) maxn = r[j];
}
len += tmp_l;
r[len++] = 0;//末尾添加一个最小值
}
da(r, sa, len, maxn+1);
calheight(r, sa, len);
int beg = 0, end = 0, ok, u = 0, ul = 0, LEN = 0;
while(left <= right)
{
ok = u = 0;
int mid = left + (right - left)/2;//二分答案
for(i=n+1; i<len; i++){
if(height[i] >= mid){//确定某一组的起点终点
for(k=sa[i]; k < sa[i] + mid; k++)
if(0 == r[k]) break;//该公共前缀含有分隔符
if(k == sa[i] + mid){
if(!beg) beg = i;
end = i;
}
}
if((beg && end) && (i == len - 1 || height[i] < mid)){
int count = 0;
MS(vis, 0);
for(j=beg-1; j<=end; j++){//一组里面有多少个后缀来自不同的字符串
int num = block[sa[j]];
if(!vis[num]) {
vis[num] = 1;
count++;
}
}
if(count > n/2){//符合题意的解
ID[u++] = sa[j-1];//保存下标
LEN = mid;
ok = 1;
}
beg = end = 0;
}
}
if(ok) ul = u;//u值在每次二分都会置为0,故在每次找到合理的解后要赋给其它变量
if(ok) left = mid + 1;//找到解，说明不是最长
else right = mid - 1;
}
if(ul){
for(i=0; i<ul; i++){
for(j=ID[i]; j<ID[i] + LEN; j++)
printf("%c", char(r[j] - 1 +'a'));
printf("\n");
}
}
else printf("?\n");
scanf("%d", &n);
if(n) printf("\n");
}
}
标签：
原文地址：http://blog.csdn.net/u013351484/article/details/43148957
