Ｎewnode's NOI(P?)模拟赛 第三题 (主席树优化建图 + tarjan)

题目/题解戳这里

这道题题目保证a,b,ca,b,ca,b,c各是一个排列…mdzz考场上想到正解但是没看到是排列,相等的情况想了半天…然后写了暴力60分走人…

由于两两间关系一定,那么就是一个竞赛图(完全图让每一条边都有向).显然就是tarjan.然后发现有很多边其实可以不存在,比如a>b>ca>b>ca>b>c,在竞赛图中就存在3条边a→b,b→c,a→ca\to b,b\to c,a\to ca→b,b→c,a→c.其实最后这一条边已经没有必要连了.那么就可以用主席树优化建边(类似BZOJ 3218 A+B problem(主席树优化网络流建图)).

CODE

选手自带常数巨大…(本地测开了O2O_2O2​还要跑2s+2s+2s+)

#include <cctype>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
inline void read(int &num) {
	char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch == '-')flg = -flg;
	for(num=0; isdigit(ch); num=num*10+ch-'0', ch=getchar()); num*=flg;
}
const int MAXN = 100005;
const int N = 100005 * 20 * 3;
int X, Y, n, m, pos[MAXN];
struct node{ int v[3], id; }p[MAXN];
int in[N], scc[N], dfn[N], stk[N], indx, tmr, scccnt;
vector<int>e[N], g[N];
inline void adde(int u, int v) { if(u&&v)e[u].push_back(v); }
inline void addg(int u, int v) { if(u&&v)g[u].push_back(v), ++in[v]; }
int tarjan(int u) {
	int lowu = dfn[u] = ++tmr;
	stk[++indx] = u;
	for(int v, i = 0, siz = e[u].size(); i < siz; ++i)
		if(!dfn[v=e[u][i]]) lowu = min(lowu, tarjan(v));
		else if(!scc[v]) lowu = min(lowu, dfn[v]);
	if(lowu == dfn[u]) { ++scccnt;
		do scc[stk[indx]] = scccnt;
		while(stk[indx--] != u);
	}
	return lowu;
}
int ls[N], rs[N];
void ins(int &i, int l, int r, int x, int id) {
	++m; ls[m] = ls[i], rs[m] = rs[i]; adde(m, i); i = m;
	if(l == r) adde(i, id);
	else {
		int mid = (l + r) >> 1;
		if(x <= mid) ins(ls[i], l, mid, x, id), adde(i, ls[i]);
		else ins(rs[i], mid+1, r, x, id), adde(i, rs[i]);
	}
}
void link(int i, int l, int r, int x, int id) {
	if(!i) return;
	if(r <= x) adde(id, i);
	else {
		int mid = (l + r) >> 1;
		link(ls[i], l, mid, x, id);
		if(x > mid) link(rs[i], mid+1, r, x, id);
	}
}
inline void work() {
	for(int i = 1; i <= n; ++i) pos[p[i].v[X]] = i;
	int root = 0;
	for(int i = 1; i <= n; ++i)
		ins(root, 1, n, p[pos[i]].v[Y], pos[i]), link(root, 1, n, p[pos[i]].v[Y], pos[i]);
}
bool flag[N]; int ans;
void topo(int u) {
	if(ans) return;
	if(flag[u]) { ans = u; return; }
	for(int i = 0, siz = g[u].size(); i < siz; ++i)
		if(--in[g[u][i]] == 0) topo(g[u][i]);
}
int main () {
	freopen("san.in", "r", stdin);
	freopen("san.out", "w", stdout);
	read(n); m = n;
	for(int i = 1; i <= n; ++i)
		for(int j = 0; j < 3; ++j)
			read(p[i].v[j]);
	X = 0, Y = 1; work();
	X = 0, Y = 2; work();
	X = 1, Y = 2; work();
	for(int i = 1; i <= m; ++i)
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= m; ++i)
		for(int j = 0, siz = e[i].size(); j < siz; ++j)
			if(scc[i] != scc[e[i][j]]) addg(scc[i], scc[e[i][j]]);
	for(int i = 1; i <= n; ++i) flag[scc[i]] = 1;
	for(int i = 1; i <= scccnt; ++i)
		if(!in[i]) topo(i);
	for(int i = 1; i <= n; ++i) puts(scc[i] == ans ? "1" : "0");
	return 0;
}