5261 Rhyme--广义SAM+拓扑排序
作者:互联网
原题链接,不是权限题
题目大意
有\(n\)个模板串,让你构造一个尽量长的串,使得这个串中任意一个长度为\(k\)的子串都是至少一个模板串的子串
题解
可以先看一下这道题 [POI2000]病毒
虽然是个\(AC\)自动机,不过思路很像
对于这道题,我们只需要把广义\(SAM\)建出来,然后在那些只经过\(maxlen\geqslant k\)的结点的路径中选一个最长的就行了。最后一步可以用拓扑排序来完成
拓扑建边时可以直接向\(fail\)连边,而不是把儿子补全(像\(AC\)自动机那样\(ch[u][c]=ch[fail[u]][c]\)),这样能降低复杂度
最后如果出现环,就输出\(INF\),否则求最长路径,注意特判所有结点的\(maxlen\)都小于\(k\)的情况,题目最下方有说明
丑的一批的代码奉上:
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define ull unsigned long long
#define pii pair<int, int>
#define mii map<int, int>
#define uint unsigned int
#define lbd lower_bound
#define ubd upper_bound
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline
#define N 100000
int n, k;
char s[N+5];
int root, nid, last, ch[2*N+5][26], fail[2*N+5], len[2*N+5];
int in[2*N+5], d[2*N+5];
vector<pii> G[2*N+5];
void clear() {
root = nid = 1;
memset(ch[1], 0, sizeof ch[1]);
memset(fail, 0, sizeof fail);
memset(in, 0, sizeof in);
memset(d, 0, sizeof d);
// memset(len, 0, sizeof len);
G[1].clear(), G[2*N+2].clear(), G[2*N+3].clear();
}
void init() {
last = root;
}
void extend(int c) {
int cur = ++nid;
memset(ch[cur], 0, sizeof ch[cur]);
G[cur].clear();
len[cur] = len[last]+1;
while(last && !ch[last][c]) ch[last][c] = cur, last = fail[last];
if(!last) fail[cur] = root;
else {
int p = last, q = ch[last][c];
if(len[q] == len[p]+1) fail[cur] = q;
else {
int clone = ++nid;
G[clone].clear();
len[clone] = len[p]+1;
for(int i = 0; i < 26; ++i) ch[clone][i] = ch[q][i];
fail[clone] = fail[q], fail[q] = fail[cur] = clone;
while(p && ch[p][c] == q) ch[p][c] = clone, p = fail[p];
}
}
last = cur;
}
int topo() {
int S = 2*N+2, T = 2*N+3, ans = 0, cnt = 0;
d[T] = 0;
G[S].clear();
for(int i = 1; i <= nid; ++i)
if(len[i] == k-1) {
for(int j = 0; j < 26; ++j) if(ch[i][j]) G[i].pb(mp(ch[i][j], 1)), in[ch[i][j]]++;
in[i]++;
G[S].pb(mp(i, len[i]));
}
else if(len[i] >= k) {
for(int j = 0; j < 26; ++j) if(ch[i][j]) G[i].pb(mp(ch[i][j], 1)), in[ch[i][j]]++;
if(fail[i] && len[fail[i]] >= k-1) G[i].pb(mp(fail[i], 0)), in[fail[i]]++;
G[i].pb(mp(T, 0));
G[S].pb(mp(i, len[i]));
in[T]++, in[i]++;
cnt++;
}
if(!cnt) return k-1;
queue<int> q;
q.push(S);
while(!q.empty()) {
int u = q.front(); q.pop();
ans = max(ans, d[u]);
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i].first, w = G[u][i].second;
in[v]--;
d[v] = max(d[v], d[u]+w);
if(!in[v]) q.push(v);
}
}
for(int i = 1; i <= nid; ++i) if(in[i]) return -1;
return ans;
}
int main() {
while(~scanf("%d%d", &n, &k)) {
clear();
for(int i = 1; i <= n; ++i) {
scanf("%s", s);
int len = strlen(s);
init();
for(int j = 0; j < len; ++j) extend(s[j]-'a');
}
int ans = topo();
if(ans == -1) printf("INF\n");
else printf("%d\n", ans);
}
return 0;
}
标签:ch,last,5261,SAM,int,Rhyme,fail,include,define 来源: https://www.cnblogs.com/dummyummy/p/10783831.html