P4316:绿豆蛙的归宿题解

主要思路

这道题是一道比较好的题(被巨佬秒切的水题)。我们在 DAG 上进行 Topo Sort 来统计答案。对于点\(i\)而言,概率为

\[ P(i) = \sum_{(u, i)} \frac{P(u)}{outDegree[u]} \]

所以期望只需要在概率上乘上边的系数,利用线性性累加即可。

代码

// P4316.cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100000 + 100;
int head[MAX_N], current, n, m, tmpx, tmpy, tmpz, indeg[MAX_N], outdeg[MAX_N];
double pro[MAX_N], ans = 0;
struct edge
{
	int to, nxt;
	double weight;
} edges[MAX_N << 1];
void addpath(int u, int v, int w)
{
	edges[current].to = v, edges[current].nxt = head[u];
	edges[current].weight = w, head[u] = current++;
}
void toposort()
{
	queue<int> q;
	q.push(1);
	pro[1] = 1;
	while (!q.empty())
	{
		int pt = q.front();
		q.pop();
		for (int i = head[pt]; i != -1; i = edges[i].nxt)
		{
			int to = edges[i].to;
			ans += (double)(pro[pt] / (double)outdeg[pt] * (double)edges[i].weight);
			pro[to] += (double)(pro[pt] / (double)outdeg[pt]);
			indeg[to]--;
			if (indeg[to] == 0)
				q.push(to);
		}
	}
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++)
		scanf("%d%d%d", &tmpx, &tmpy, &tmpz), addpath(tmpx, tmpy, tmpz), indeg[tmpy]++, outdeg[tmpx]++;
	toposort();
	printf("%.2lf", ans);
	return 0;
}

发布者

Kal0rona

江西师大附中全机房最弱

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