【LOJ6403】「ICPC World Finals 2018」赶飞机

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/88657750

【题目链接】

• 点击打开链接

【思路要点】

• 将每一辆公交车的出发站，出发时间、目标站，到达时间视为关键点，按照时间从大到小 $dp$ 即可。
• 时间复杂度 $O(MLogM)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e6 + 5;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int n, m, tot, pos[MAXN]; ll k;
map <pair <ll, int>, int> home;
pair <ll, int> x[MAXN], y[MAXN];
double ans[MAXN], dp[MAXN], p[MAXN];
vector <pair <int, double> > a[MAXN];
int main() {
	read(m), read(n), read(k);
	static set <pair <ll, int>> st;
	for (int i = 1; i <= m; i++) {
		read(x[i].second), read(y[i].second);
		read(x[i].first), read(y[i].first), y[i].first++;
		st.insert(x[i]), st.insert(y[i]);
		scanf("%lf", &p[i]);
	}
	for (auto x : st) {
		pos[++tot] = x.second;
		home[x] = tot;
	}
	for (int i = 1; i <= m; i++)
		a[home[x[i]]].emplace_back(home[y[i]], p[i]);
	ans[1] = 1;
	for (int i = tot; i >= 1; i--) {
		dp[i] = ans[pos[i]];
		for (auto x : a[i])
			chkmax(dp[i], ans[pos[i]] * (1 - x.second) + dp[x.first] * x.second);
		ans[pos[i]] = dp[i];
	}
	printf("%.10lf\n", ans[0]);
	return 0;
}