BZOJ4567 [Scoi2016]背单词 【trie树 + 贪心】

题目链接

BZOJ4567

题解

题意真是鬼畜= =
意思就是说我们应先将一个串的所有后缀都插入之后再插入这个串，产生代价为其到上一个后缀的距离

我们翻转一下串，转化为前缀，就可以建(trie)树来解决了
建好(trie)后单独取出单词节点，贪心先往子树小的节点编号即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 600005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
char s[maxm];
int n,ch[maxm][26],fail[maxm],last[maxm],tag[maxm],cnt;
void ins(int x){
	int u = 0,len = strlen(s + 1),id;
	reverse(s + 1,s + 1 + len);
	for (int i = 1; i <= len; i++){
		id = s[i] - 'a';
		u = ch[u][id] ? ch[u][id] : (ch[u][id] = ++cnt);
	}
	tag[u] = x;
}
LL ans;
int now,id[maxm],fa[maxm],siz[maxm];
priority_queue<cp,vector<cp>,greater<cp> > son[maxm];
void dfs1(int u){
	siz[u] = tag[u] > 0;
	for (int i = 0; i < 26; i++){
		if (!ch[u][i]) continue;
		fa[ch[u][i]] = tag[u] ? u : fa[u];
		dfs1(ch[u][i]);
		siz[u] += siz[ch[u][i]];
	}
	if (tag[u]) son[fa[u]].push(mp(siz[u],u));
}
void dfs2(int u){
	if (tag[u]){
		id[u] = ++now;
		ans += id[u] - id[fa[u]];
	}
	while (!son[u].empty()){
		int to = son[u].top().second; son[u].pop();
		dfs2(to);
	}
}
int main(){
	n = read();
	REP(i,n){
		scanf("%s",s + 1);
		ins(i);
	}
	dfs1(0);
	dfs2(0);
	printf("%lld
",ans);
	return 0;
}