SUBLEX Lexicographical Substring Search 求字典序第k小的子串
阿新 • • 發佈:2019-02-20
題目:求字串字典序第k小的子串
思路:統計每個狀態的子串的個數,按字典序尋找
程式碼:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<algorithm> #include<ctime> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<vector> #include<map> #include<set> #include<queue> #include<stack> #include<list> #include<numeric> using namespace std; #define LL long long #define ULL unsigned long long #define INF 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define PP puts("*********************"); template<class T> T f_abs(T a){ return a > 0 ? a : -a; } template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; } template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} // 0x3f3f3f3f3f3f3f3f //0x3f3f3f3f const int MAXN = 200050, SIZE = 26; struct SAM { int len[MAXN], link[MAXN], next[MAXN][SIZE]; int total, last; inline int newNode(int L) { len[++total] = L; link[total] = 0; for(int i = 0; i < SIZE; ++i) next[total][i] = 0; return total; } inline void Add(int c) { int i, p = last, cur = newNode(len[last] + 1); for(; p && !next[p][c]; p = link[p]) next[p][c] = cur; if(!p) link[cur] = 1;//令其指向初始狀態 else { int q = next[p][c]; if(len[q] == len[p] + 1) link[cur] = q; else {//> int clone = newNode(len[p] + 1); for(i = 0; i < SIZE; ++i) next[clone][i] = next[q][i]; link[clone] = link[q]; link[q] = link[cur] = clone; for(; p && next[p][c] == q; p = link[p]) next[p][c] = clone; } } last = cur; } void Init () {//根節點是1 total = 0; last = newNode(0); } }sam; char str[MAXN]; int num[MAXN],idx[MAXN]; int dp[MAXN]; void solve(int k){ int now=1,L=0; while(k){ for(int j=0;j<SIZE;j++){ if(sam.next[now][j]==0) continue; int x=sam.next[now][j]; if(k>dp[x]) k-=dp[x]; else{ str[L++]='a'+j; now=x; k--; break; } } } str[L]='\0'; puts(str); } int main(){ int Q,k; scanf("%s",str); sam.Init(); for(int i=0;str[i]!='\0';i++) sam.Add(str[i]-'a'); for(int i=1;i<=sam.total;i++) num[i]=0; for(int i=1;i<=sam.total;i++) num[sam.len[i]]++; for(int i=1;i<=sam.total;i++) num[i]+=num[i-1]; for(int i=1;i<=sam.total;i++) idx[num[sam.len[i]]--]=i; for(int i=sam.total;i>=1;i--){ int now=idx[i]; dp[now]=1; for(int j=0;j<SIZE;j++) dp[now]+=dp[sam.next[now][j]]; } scanf("%d",&Q); while(Q--){ scanf("%d",&k); solve(k); } return 0; }