【BZOJ4481】【JSOI2015】非诚勿扰

【题目链接】

• 点击打开链接
  

【思路要点】

• 用等比数列求和公式求出每条边出现的概率。
• 对于每一条边，用树状数组统计与它交叉的边的出现概率和，计算答案即可。
• 时间复杂度\(O(NLogN)\)（\(N\)，\(M\)同阶）。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 500005;
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("");
}
struct BinaryIndexTree {
	long double a[MAXN]; int n;
	void init(int x) {
		n = x;
		memset(a, 0, sizeof(a));
	}
	void modify(int x, long double p) {
		x = n - x + 1;
		for (int i = x; i <= n; i += i & -i)
			a[i] += p;
	}
	long double query(int x) {
		x = n - x;
		long double ans = 0;
		for (int i = x; i >= 1; i -= i & -i)
			ans += a[i];
		return ans;
	}
} BIT;
int n, m;
long double p;
vector <int> a[MAXN];
int main() {
	read(n), read(m);
	scanf("%Lf", &p);
	for (int i = 1; i <= m; i++) {
		int x, y;
		read(x), read(y);
		a[x].push_back(y);
	}
	BIT.init(n);
	long double ans = 0;
	for (int i = 1; i <= n; i++) {
		sort(a[i].begin(), a[i].end());
		long double r = pow(1 - p, a[i].size()), x = p;
		for (unsigned j = 0; j < a[i].size(); j++) {
			long double tmp = x / (1 - r);
			ans += tmp * BIT.query(a[i][j]);
			BIT.modify(a[i][j], tmp);
			x *= 1 - p;
		}
	}
	printf("%.2Lf\n", ans);
	return 0;
}