按照以前两个字符串找两者的最长公共子串的思路类似，可以把所有串拼接到一起，这里为了避免讨论LCP跨越多个串需需要特别处理的问题用不同的字符把所有串隔开（因为char只有128位，和可能不够用，更推荐设置成某一特殊字符，在后缀数组初始化的时候在对其映射的int值做处理）。二分长度然后遍历Height，判断SA是否处于同一个串。这里我一开始觉得判断不同串的计数有点麻烦，其实根据LCP确定起点，随终点延伸Height值不会上升的特性，一旦Height小于我们要求的值就切换到下一计数（因为必然不可能再和之前计数的串再产生重合了）。
#include <iostream> #include <cstring> #include <string> #include <cmath> #include <vector> #include <algorithm> #include <cstdio> #define debug puts("debug") #define LL long long using namespace std; const int N=111005; string s,temp; int nn; int pos[N]; int vis[N]; int phs[N]; int endp[N]; int stx[N],sp; class SF { //N:数组大小 public: int x[N], y[N], c[N]; int Height[N], str[N], SA[N], Rank[N];//Height数组从2开始,SA记录Rank=i的下标 int slen; int m;//字符集处理大小(传入如果不是数字，需要做位移转换) bool cmp(int* r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } void Suffix(int n) { ++n; int i, j, p; for (i = 0; i < m; ++i) c[i] = 0; for (i = 0; i < n; ++i) c[x[i] = str[i]]++; for (i = 1; i < m; ++i) c[i] += c[i - 1]; for (i = n - 1; i >= 0; --i) SA[--c[x[i]]] = i; for (j = 1; j <= n; j <<= 1) { p = 0; for (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 < m; ++i) c[i] = 0; for (i = 0; i < n; ++i) c[x[y[i]]]++; for (i = 1; i < m; ++i) c[i] += c[i - 1]; for (i = n - 1; i >= 0; --i) SA[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[SA[0]] = 0; for (i = 1; i < n; ++i) { x[SA[i]] = cmp(y, SA[i - 1], SA[i], j) ? p - 1 : p++; } if (p >= n)break; m = p; } int k = 0; n--; for (i = 0; i <= n; ++i) Rank[SA[i]] = i; for (i = 0; i < n; ++i) { if (k)--k; j = SA[Rank[i] - 1]; while (str[i + k] == str[j + k])++k; Height[Rank[i]] = k; //cout << k << endl; } } static const int bitlen = 35; LL lg2(LL p)//计算log2(n) { return (LL)(log(p) / log(2)); } LL dp[bitlen][N]; LL bit[bitlen]; void initRMQ()//初始化 { bit[0] = 1; for (int i = 1; i < bitlen; i++) bit[i] = 2 * bit[i - 1]; for (int i = 0; i <= slen; i++) dp[0][i] = Height[i]; dp[0][0] = dp[0][1] = 0; for (LL i = 1; bit[i] < slen + 1; i++) for (LL j = 0; j + bit[i] <= slen + 1; j++) dp[i][j] = min(dp[i - 1][j], dp[i - 1][j + bit[i - 1]]); } LL query(LL l, LL r)//查询两个Rank之间的lcp { if (r == l) return slen - SA[l]; if (l > r) swap(l, r); l++; LL mig = lg2(r - l + 1.0); return min(dp[mig][l], dp[mig][r - bit[mig] + 1]); } void init(string s) { slen = s.size(); m=1050;//每次都需要初始化m int fuck=1; for (int i = 0; i < slen; i++) { if(s[i]!='#') str[i] = s[i] - 'a' + 2;//如果是字符，映射成从1开始的序列 else str[i]=100+(fuck++); } str[slen] = 0;//1作为结束符，防止越界 Suffix(slen); initRMQ(); } bool ok(int lx) { sp=0; fill(vis,vis+slen+1,0); int cnt=1; vis[phs[SA[1]]]=1; //cout<<phs[1]<<endl; for(int i=2; i<=slen; i++) { //cout<<Height[i]<<endl; //cout<<phs[i]<<endl; if(Height[i]>=lx) { if(vis[phs[SA[i]]]==0) cnt++,vis[phs[SA[i]]]=1; } else { if(cnt>nn/2) stx[sp++]=SA[i-1]; cnt=1; fill(vis,vis+nn+1,0); vis[phs[SA[i]]]=1; } } if(cnt>nn/2) stx[sp++]=SA[slen-1]; return sp; } int bins(int l,int r) { while(r-l>=3) { int mid=(l+r)/2; if(ok(mid))l=mid; else r=mid-1; } //cout<<l<<' '<<r<<endl; for(; r>=l; r--) { //cout<<r<<endl; if(ok(r))return r; } return 0; } void solve() { int f=bins(1,slen); if(!f)cout<<"?"<<endl; else { //sort(stx,stx+f,cmp); for(int i=0; i<sp; i++) { //cout<<stx[i]<<endl; for(int j=0; j<f; j++) cout<<s[stx[i]+j]; cout<<endl; } } } } sf; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); cin.sync_with_stdio(false); int cnt=0; while(cin>>nn) { if(!nn)break; if(cnt++)cout<<endl; s=""; for(int i=0; i<nn; i++) { cin>>temp; pos[i]=temp.length(); s+=temp; s+='#'; } if(nn==1) { s=s.substr(0,s.length()-1); cout<<s<<endl; continue; } int p=0; for(int i=0; i<nn; i++) { for(int j=0; j<pos[i]; j++,p++) phs[p]=i; phs[p++]=nn; if(i==0)endp[i]=pos[i]; else endp[i]=endp[i-1]+pos[i]+1; } sf.init(s); sf.solve(); //sf.ok(1); } return 0; }