ACdream 1430 SETI 后缀自动机

题目大意：

就是对于给定的串S, 长度不超过10000, 求出其有多少种子串T, 使得T在S中至少不重叠地出现了2次以上

大致思路：

很明显的后缀自动机吧, 建立S的后缀自动机, 记录每个节点对应的串出现的最左和最右位置, 以及出现次数, dfs遍历一遍后缀自动机上的点即可了

细节见代码吧

代码如下：

Result  :  Accepted     Memory  :  7076 KB     Time  :  8 ms

/*
* Author: Gatevin
* Created Time: 2015/5/20 20:29:44
* 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 20010
#define maxm 10010

struct Suffix_Automation
{
struct State
{
State *par;
State *go[26];
int right, val, leftmost, rightmost, mi;
bool vis;
void init(int _val = 0)
{
par = 0, leftmost = 1e9 + 7, rightmost = -1, val = _val, right = 0, mi = 0;
vis = 0;
memset(go, 0, sizeof(go));
}
};
State *root, *last, *cur;
State nodePool[maxn];
void init()
{
cur = nodePool;
root = newState();
last = root;
}
State* newState(int val = 0)
{
cur->init(val);
return cur++;
}
void extend(int w, int pos)
{
State *p = last;
State *np = newState(p->val + 1);
np->right = 1;
np->leftmost = pos, np->rightmost = pos;
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;
}
State* b[maxn];
int d[maxm];
void topo()
{
int cnt = cur - nodePool;
int maxVal = 0;
memset(d, 0, sizeof(d));
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 info()
{
int cnt = cur - nodePool;
State *p;
for(int i = cnt - 1; i > 0; i--)
{
p = b[i];
p->par->right += p->right;
p->mi = p->par->val + 1;
p->par->leftmost = min(p->par->leftmost, p->leftmost);
p->par->rightmost = max(p->par->rightmost, p->rightmost);
}
}
int ans;
void dfs(State *now)
{
now->vis = 1;
if(now->right >= 2 && now != root)
{
int len2 = now->rightmost - now->leftmost;
if(now->val <= len2) ans += now->val - now->mi + 1;
else if(now->mi <= len2) ans += len2 - now->mi + 1;
}
for(int i = 0; i < 26; i++)
if(now->go[i] && !now->go[i]->vis)
dfs(now->go[i]);
return;
}
void solve()
{
ans = 0;
dfs(root);
printf("%d\n", ans);
}
};

Suffix_Automation sam;
char in[10010];

int main()
{
while(scanf("%s", in) != EOF)
{
int len = strlen(in);
sam.init();
for(int i = 0; i < len; i++)
sam.extend(in[i] - 'a', i);
sam.topo();
sam.info();
sam.solve();
}
return 0;
}