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Luogu1081 开车旅行

时间:2018-10-29 10:28:55      阅读:195      评论:0      收藏:0      [点我收藏+]

标签:处理   algo   一个   for   nbsp   pair   insert   pos   def   

代码巨长的倍增题...

显然这是没有什么决策的,选择方案都是固定的

这可以考虑倍增跳,每个位置上跳到的位置是可以通过查前驱后继解决的

有两种方式:

  一种是平衡树搞法,倒着做查完了插入

  另一种是先排序建一个链表,查完了删除

都是可以通过加哨兵节点来搞的,
结果我只想到了 set 乱搞,就写了很长

预处理完就可做了

第一问对于每个点倍增一遍,其实就是照题意模拟,倍增优化一下

第二问还是照题意模拟,倍增优化一下

暴力有 70pts ?


代码:

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cstdio>
#include <locale>
#include <cmath>
#include <set>
using namespace std;

typedef long long ll;
const int MAX_N = 100005;
const double INF = 100000000000001.0;

struct Node {
	int len, id;
	Node(int L = 0, int ID = 0) {len = L; id = ID;}
	bool operator < (const Node& b) const {
		return len < b.len;
	}
};
struct PAIR {
	ll fir, sec;
	PAIR(ll x = 0, ll y = 0) {fir = x; sec = y;}
	friend PAIR operator + (PAIR a, PAIR b) {
		return PAIR(a.fir + b.fir, a.sec + b.sec);
	}
}d[MAX_N][20][2];
int n, lg, x0, m, ans;

int h[MAX_N], f[MAX_N][20][2];
double ans_rat;
set<Node> st;

inline int rd() {
	register int x = 0, c = getchar();
	register bool f = false;
	while (!isdigit(c)) {
		f = (c == ‘-‘);
		c = getchar();
	}
	while (isdigit(c)) {
		x = x * 10 + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}
inline void get_for(int pos, set<Node>::iterator iter) {
	auto lef = iter, rig = iter;
	if (iter == st.begin()) {
		f[pos][0][0] = iter->id;
		d[pos][0][0].fir = abs(h[pos] - h[iter->id]);
		++iter;
		if (st.size() == 1) return;
		f[pos][0][1] = iter->id;
		d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
		return;
	} else if (iter == st.end()) {
		--iter;
		f[pos][0][0] = iter->id;
		d[pos][0][0].fir = abs(h[pos] - h[iter->id]);
		if (iter == st.begin()) return;
		--iter;
		f[pos][0][1] = iter->id;
		d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
		return;
	}
	++rig;
	if (rig == st.end()) {
		--lef;
		if (lef == st.begin()) {
			if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) {
				f[pos][0][0] = lef->id;
				d[pos][0][0].fir = abs(h[pos] - h[lef->id]);
				f[pos][0][1] = iter->id;
				d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
			} else {
				f[pos][0][0] = iter->id;
				d[pos][0][0].fir = abs(h[pos] - h[iter->id]);
				f[pos][0][1] = lef->id;
				d[pos][0][1].sec = abs(h[pos] - h[lef->id]);
			}
			return;
		}
		if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) {
			f[pos][0][0] = lef->id;
			d[pos][0][0].fir = abs(h[pos] - h[lef->id]);
			--lef;
			if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) {
				f[pos][0][1] = lef->id;
				d[pos][0][1].sec = abs(h[pos] - h[lef->id]);
			} else {
				f[pos][0][1] = iter->id;
				d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
			}
			return;
		}
		f[pos][0][0] = iter->id;
		d[pos][0][0].fir = abs(h[pos] - h[iter->id]);
		f[pos][0][1] = lef->id;
		d[pos][0][1].sec = abs(h[pos] - h[lef->id]);
		return;
	}
	--lef;
	if (abs(h[pos] - h[lef->id]) == abs(h[pos] - h[iter->id])) {
		f[pos][0][0] = lef->id;
		d[pos][0][0].fir = abs(h[pos] - h[lef->id]);
		f[pos][0][1] = iter->id;
		d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
	} else if (abs(h[pos] - h[lef->id]) < abs(h[pos] - h[iter->id])) {
		f[pos][0][0] = lef->id;
		d[pos][0][0].fir = abs(h[pos] - h[lef->id]);
		if (lef == st.begin()) {
			f[pos][0][1] = iter->id;
			d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
		} else {
			--lef;
			if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) {
				f[pos][0][1] = lef->id;
				d[pos][0][1].sec = abs(h[pos] - h[lef->id]);
			} else {
				f[pos][0][1] = iter->id;
				d[pos][0][1].sec = abs(h[pos] - h[iter->id]);
			}
		}
	} else {
		f[pos][0][0] = iter->id;
		d[pos][0][0].fir = abs(h[pos] - h[iter->id]);
		if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[rig->id])) {
			f[pos][0][1] = lef->id;
			d[pos][0][1].sec = abs(h[pos] - h[lef->id]);
		} else {
			f[pos][0][1] = rig->id;
			d[pos][0][1].sec = abs(h[pos] - h[rig->id]);
		}
	}
}
inline void DBL_init() {
	for (int i = n; i >= 1; --i) {
		if (!st.empty()) {
			auto dst = st.lower_bound(h[i]);
			get_for(i, dst);
		}
		st.insert(Node(h[i], i));
	}
	for (int i = 1; i <= n && 2 + i <= n; ++i) {
		f[i][1][0] = f[f[i][0][0]][0][1];
		f[i][1][1] = f[f[i][0][1]][0][0];
		d[i][1][0] = d[i][0][0] + d[f[i][0][0]][0][1];
		d[i][1][1] = d[i][0][1] + d[f[i][0][1]][0][0];
	}
	for (int j = 2; j <= lg; ++j) {
		for (int i = 1; i <= n && (1 << j) + i <= n; ++i) {
			f[i][j][0] = f[f[i][j - 1][0]][j - 1][0];
			f[i][j][1] = f[f[i][j - 1][1]][j - 1][1];
			d[i][j][0] = d[i][j - 1][0] + d[f[i][j - 1][0]][j - 1][0];
			d[i][j][1] = d[i][j - 1][1] + d[f[i][j - 1][1]][j - 1][1];
		}
	}
}
inline void get_ans(int bgn) {
	register int pos = bgn;
	register ll dst_a = 0, dst_b = 0;
	register bool got_bgn = false;
	register double rat;
	for (int i = lg; i >= 0; --i) {
		if (!f[pos][i][1] || dst_a + dst_b + d[pos][i][1].fir + d[pos][i][1].sec > x0 || (i == 0 && !f[pos][0][0])) continue;
		got_bgn = true;
		dst_b += d[pos][i][1].fir;
		dst_a += d[pos][i][1].sec;
		pos = f[pos][i][1];
	}
	if (!got_bgn) return;
	if (dst_b == 0) {
		if (dst_a == 0) rat = 1.0;
		else rat = INF;
	} else {
		rat = double(dst_a) / double(dst_b);
	}
	if (fabs(0.0 - ans_rat) < 1e-7 || rat < ans_rat || (fabs(rat - ans_rat) < 1e-7 && h[bgn] > h[ans])) {
		ans_rat = rat;
		ans = bgn;
	}
}

int main() {
	n = rd();
	lg = int(ceil(log2(n)));
	for (int i = 1; i <= n; ++i) h[i] = rd();
	DBL_init();
	x0 = rd();
	for (int i = 1; i <= n; ++i)
		get_ans(i);
	printf("%d\n", ans);
	m = rd();
	register int s = 0, x = 0;
	register ll tot_a = 0, tot_b = 0;
	while (m--) {
		tot_a = tot_b = 0;
		s = rd(); x = rd();
		for (int i = lg; i >= 0; --i) {
			if (!f[s][i][1] || (tot_a + tot_b + d[s][i][1].fir + d[s][i][1].sec > x)) continue;
			tot_b += d[s][i][1].fir;
			tot_a += d[s][i][1].sec;
			s = f[s][i][1];
		}
		printf("%lld %lld\n", tot_a, tot_b);
	}
	return 0;
}

 

Luogu1081 开车旅行

标签:处理   algo   一个   for   nbsp   pair   insert   pos   def   

原文地址:https://www.cnblogs.com/xcysblog/p/9868559.html

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