nowcoder 207G - Coding Contest - [最小费用最大流]

Posted 2021-01-11 dilthey

题目链接：https://www.nowcoder.com/acm/contest/207/G

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the ui-th block to the vi-th block. Your task is to solve the lunch issue. According to the arrangement, there are si competitors in the i-th block. Limited to the size of table, bi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path.
For the i-th path, the wires have been stabilized at ?rst and the ?rst competitor who walker through it would not break the wires. Since then, however, when a person go through the i?th path, there is a chance of pi to touch the wires and a?ect the whole networks. Moreover, to protect these wires, no more than ci competitors are allowed to walk through the i-th path.
Now you need to ?nd a way for all competitors to get their lunch, and minimize the possibility of network crashing.
输入描述:
The ?rst line of input contains an integer t which is the number of test cases. Then t test cases follow.
For each test case, the ?rst line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and bi (si,bi ≤ 200).
Each of the next M lines contains three integers ui,vi and ci(ci ≤ 100) and a ?oat-point number pi(0 < pi < 1). It is guaranteed that there is at least one way to let every competitor has lunch.
输出描述:
For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.
示例1
输入
1
4 4
2 0
0 3
3 0
0 3
1 2 5 0.5
3 2 5 0.5
1 4 5 0.5
3 4 5 0.5
输出
0.50

 

题意：

题目的灵感估计来源于现场赛发餐包的操作……

现在赛场划分成了 $N$ 个不相交区域，共有 $M$ 条有向路连接两个区域，

对于每个区域，给出 $s[i],b[i]$ 代表区域内有 $s[i]$ 个人，$b[i]$ 个餐包，一旦某人在本区域内拿不到餐包，就会前往其他区域获取餐包，

而众所周知，赛场上的路上是有很多电线的，一不小心就会踢到电线，所以现在每条路上都存在这一些电线，

现在已知，一旦某个选手走过第 $i$ 条有向边，就有 $p[i]$ 的概率踢到电线，进而影响整个电网，不过经过该路径的第一个人是必然不会踢到电线的，同时对于第 $i$ 条边，限制最多 $c[i]$ 个人走过。

现在，求整个电网被影响的最小概率。

 

题解：

首先，我们考虑既然将来还要深入学习数学相关知识，概率和期望这一块是怎么样都跑不掉的，所以还不如现在趁机好好巩固一下概率论的基础知识……

考虑每个人踢到电线的概率是相互独立的，我们将某次某个人经过某条边称作一次实验，

若每次实验踢到电线事件发生的概率相同，则 $n$ 个人踢到电线 $k$ 次的概率服从二项分布，众所周知二项分布的公式为

$Pleft( {X = k} ight) = C_n^k p^k left( {1 - p} ight)^{n - k}$

其中 $p$ 即为一次实验中发生踢到电线事件的概率；

而发生踢到电线这一事件发生 $1,2,3,cdots$ 次均会影响电网，所以总共进行 $n$ 次独立实验后，电网被影响的概率为

$sumlimits_{k = 1}^n {Pleft( {X = k} ight)} = 1 - Pleft( {X = 0} ight)$

易知

$Pleft( {X = 0} ight) = left( {1 - p} ight)^n$

但是我们知道，电网被影响的概率为

$Pleft( {X = 1,2, cdots ,n} ight) = 1 - left( {1 - p} ight)^n$

当然，本题中，每次实验踢电线事件发生概率不一定相同，但是概率的计算方法依然符合上式：

$P = 1 - prodlimits_{i = 1}^n {left( {1 - pleft[ i ight]} ight)}$

也就是说，