SPOJ SUBLEX Lexicographical Substring Search 后缀自动机

题目大意：

就是现在对于给出的长度不超过90000的字符串进行Q(Q <= 500)次询问, 每次询问给出一个K， 要求输出第K小的子串(0 < K < 2^31)

其中相同的子串只算一次例如“aaa"的子串是"a", "aa", "aaa"

大致思路：

首先对于给出的字符串建立后缀自动机, 然后利用后缀自动机的性质, 所有相同的子串一定会在同一点终止, 那么, 从根开始, 每次都选择尽量小的字符走, 首先我们可以dfs预处理出每个状态点处代表的可能向下的不同字串有多少个, 然后就知道每个字符向下走会有多少种可能了, 于是根据这个图中每次沿着满足向下能找到第K小的边走即可

代码如下：

Result  :  Accepted     Memory  :  28672 KB     Time  :  410 ms

/*
* Author: Gatevin
* Created Time: 2015/4/10 9:54:27
* File Name: Rin_Tohsaka.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)
#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl

#define maxn 90010*2
#define maxm 90010

struct Suffix_Automation
{
struct State
{
State *par;
State *go[26];
int val, mi, right;
lint cnt;
void init(int _val = 0)
{
par = 0, val = _val, mi = 0, cnt = 0, right = 0;
memset(go, 0, sizeof(go));
}
int calc()
{
if(par == 0) return 0;
return val - par->val;
}
};
State *root, *last, *cur;
State nodePool[maxn];
State* newState(int val = 0)
{
cur->init(val);
return cur++;
}
void initSAM()
{
cur = nodePool;
root = newState();
last = root;
}
void extend(int w)
{
State *p = last;
State *np = newState(p->val + 1);
np->right = 1;
while(p && p->go[w] == 0)
{
p->go[w] = np;
p = p->par;
}
if(p == 0)
{
np->par = root;
}
else
{
State *q = p->go[w];
if(p->val + 1 == q->val)
{
np->par = q;
}
else
{
State *nq = newState(p->val + 1);
memcpy(nq->go, q->go, sizeof(q->go));
nq->par = q->par;
q->par = nq;
np->par = nq;
while(p && p->go[w] == q)
{
p->go[w] = nq;
p = p->par;
}
}
}
last = np;
}

int d[maxm];
State* b[maxn];
void topo()
{
int cnt = cur - nodePool;
memset(d, 0, sizeof(d));
int maxVal = 0;
for(int i = 1 ; i < cnt; i++)
maxVal = max(maxVal, nodePool[i].val), d[nodePool[i].val]++;
for(int i = 1; i <= maxVal; i++) d[i] += d[i - 1];
for(int i = 1; i < cnt; i++) b[d[nodePool[i].val]--] = &nodePool[i];
b[0] = root;
}

void SAMInfo()
{
State *p;
int cnt = cur - nodePool;
for(int i = cnt - 1; i > 0; i--)
{
p = b[i];
p->par->right += p->right;
p->mi = p->par->val + 1;
}
}

lint dfs(State *now)//返回从now状态向下有多少种可能的串
{
if(now->cnt) return now->cnt;
lint ret = 0;
for(int i = 0; i < 26; i++)
if(now->go[i])
ret += dfs(now->go[i]);
return now->cnt = ret + (now != root);
}

void solve(lint K)//output the Kth smallest substring
{
State *now = root;
while(K)
{
for(int i = 0; i < 26; i++)
{
if(!now->go[i]) continue;
if(K > now->go[i]->cnt)
K -= now->go[i]->cnt;
else
{
putchar(i + 'a');
K--;
now = now->go[i];
break;
}
}
}
putchar('\n');
return;
}
};

Suffix_Automation sam;
char s[maxm];
int Q, K;

int main()
{
scanf("%s", s);
int len = strlen(s);
sam.initSAM();
for(int i = 0; i < len; i++)
sam.extend(s[i] - 'a');
sam.topo();
sam.SAMInfo();
sam.dfs(sam.root);
scanf("%d", &Q);
while(Q--)
{
scanf("%d", &K);
sam.solve((lint)K);
}
return 0;
}