bzoj 3073: [Pa2011]Journeys

题目:

题目(别找我要权限号，我没有)

这一题裸的线段树优化建图，直接码就是了。我尝试一下不看其他人的代码，直接打代码，结果险些没有过，加了快读，并且时限20s，空间512MB才过(处理器i3-6100)(bzoj的题目上面写的就是这个标准)。

果然我还是太弱了。

Code:

//线段树优化建图
//函数名不要在意
//函数返回的void不要在意
//Chtolly
#include <bits/stdc++.h>
#ifdef wll
#include <windows.h>
#endif
#define xx first
#define yy second
#define ll long long
#define _for(i,a,b) for(int i=(a);i<=(b);++i)
#define abs(n) ((n) > 0 ? (n) : -(n))

using namespace std;

struct _in {
	const _in &operator, (int &a) const {
		a = 0;
		char k = getchar ();
		int f = 1;
		while (k > '9' || k < '0') {
			if (k == '-') {
				f = -1;
			}
			k = getchar ();
		}
		while (k >= '0' && k <= '9') {
			a = a * 10 + k - '0';
			k = getchar ();
		}
		a *= f;
		return *this;
	}
}in;

const int N = 1000000 + 5, M = N * 10;
int l[M], r[M], ch[M][2], tot;
int fr[M << 1], to[M << 1], h[M], tot2;
bool w[M];

inline void add (int u, int v, bool l) {
	tot2++;
	fr[tot2] = h[u];
	to[tot2] = v;
	h[u] = tot2;
	w[tot2] = l;
	return (void)("神奇");
}

int n, m, P, root1, root2;
int storer[M], Pos[M];

void build (int u, int v, bool con) {
	int p = ++tot;
	l[p] = u;
	r[p] = v;
	if (u == v) {
		if (con) {
			storer[u] = p;
		}
		else {
			add (storer[u], p, 0);
			Pos[u] = p;
		}
		return (void)("毒瘤");
	}
	int mid = (u + v) / 2;
	if (con) {
		add (p, tot + 1, 0);
	}
	else {
		add (tot + 1, p, 0);
	}
	ch[p][0] = tot + 1;
	build (u, mid, con);
	if (con) {
		add (p, tot + 1, 0);
	}
	else {
		add (tot + 1, p, 0);
	}
	ch[p][1] = tot + 1;
	build (mid + 1, v, con);
	return (void)("真的毒瘤");
}

void solve (int u, int v, int p, int aim, bool con) {
	if (u <= l[p] && v >= r[p]) {
		if (con) {
			add (aim, p, 0);
		}
		else {
			add (p, aim, 0);
		}
		return ;
	}
	int mid = (l[p] + r[p]) / 2;
	if (u <= mid) {
		solve (u, v, ch[p][0], aim, con);
	}
	if (v > mid) {
		solve (u, v, ch[p][1], aim, con);
	}
	return (void)("OI大法好");
}

inline void ak (int u, int v, int m, int n) {
	tot += 2;
	solve (u, v, root2, tot - 1, 0);
	add (tot - 1, tot, 1);
	solve (m, n, root1, tot, 1);
	return (void)("退役了");
}

int dis[M];
bool inq[M];

inline void AK() {
	int f = Pos[P];
	memset (dis, 0x7F, sizeof (dis));
	dis[f] = 0;
	priority_queue <pair<int, int> > Q;
	Q.push (make_pair (0, f));
	inq[f] = 1;
	while (Q.size()) {
		int u = Q.top().second;
		Q.pop ();
		inq[u] = 0;
		for (int i = h[u]; i; i = fr[i]) {
			if (dis[to[i]] > dis[u] + w[i]) {
				dis[to[i]] = dis[u] + w[i];
				if (!inq[to[i]]) {
					inq[to[i]] = 1;
					Q.push(make_pair (-dis[to[i]], to[i]));
				}
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		printf ("%d\n", dis[Pos[i]]);
	}
	return (void)("快A");
}

int main () {
//	#ifdef wll
	freopen ("testdata.in", "r", stdin);
	freopen ("testdata.out", "w", stdout);
//	#endif
	scanf ("%d%d%d", &n, &m, &P);
	root1 = tot + 1;
	build (1, n, 1);
	root2 = tot + 1;
	build (1, n, 0);
	for (int i = 1; i <= m; ++i) {
		int a, b, c, d;
		in, a, b, c, d;
		ak (a, b, c, d);
		ak (c, d, a, b);
	}
	AK();
	return 0 + 0 - 0;
}