hdu-5988 Coding Contest(费用流)
题目链接:
Coding Contest
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
For the i-th path, the wires have been stabilized at first and the first 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 affect 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 find a way for all competitors to get their lunch, and minimize the possibility of network crashing.
For each test case, the first 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 float-point number pi(0 < pi < 1).
It is guaranteed that there is at least one way to let every competitor has lunch.
#include <bits/stdc++.h>
using namespace std;
const int maxn=500;
const double inf=1e9;
const double eps=1e-8;
struct Edge
{
int from,to,cap,flow;
double cost;
};
int n,m,s,t,M;
std::vector<int> G[maxn];
std::vector<Edge> edge;
int inq[maxn],a[maxn],p[maxn];
double d[maxn];
inline void add_edge(int from,int to,int cap,double cost)
{
edge.push_back((Edge){from,to,cap,0,cost});
edge.push_back((Edge){to,from,0,0,-cost});
m=edge.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool bellmanford(int &flow,double &cost)
{
for(int i=0;i<=t+1;i++)d[i]=inf;
memset(inq,0,sizeof(inq));
d[s]=0;inq[s]=1;p[s]=0;a[s]=inf;
queue<int>qu;
qu.push(s);
while(!qu.empty())
{
int fr=qu.front();qu.pop();
inq[fr]=0;
int len=G[fr].size();
for(int i=0;i<len;i++)
{
Edge& e=edge[G[fr][i]];
if(e.cap>e.flow&&d[e.to]>d[fr]+e.cost+eps)
{
d[e.to]=d[fr]+e.cost;
p[e.to]=G[fr][i];
a[e.to]=min(a[fr],e.cap-e.flow);
if(!inq[e.to]){qu.push(e.to);inq[e.to]=1;}
}
}
}
if(d[t]>=inf)return false;
flow+=a[t];
cost+=d[t]*a[t];
int u=t;
while(u!=s)
{
edge[p[u]].flow+=a[t];
edge[p[u]^1].flow-=a[t];
u=edge[p[u]].from;
}
return true;
}
double mincostflow()
{
int flow=0;double cost=0;
while(bellmanford(flow,cost));
return cost;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int u,v,w;
double xp;
scanf("%d%d",&n,&M);
s=0,t=n+1;
edge.clear();
for(int i=0;i<=t;i++)G[i].clear();
for(int i=1;i<=n;i++)
{
scanf("%d%d",&u,&v);
add_edge(s,i,u,0.0);
add_edge(i,t,v,0.0);
}
for(int i=1;i<=M;i++)
{
scanf("%d%d%d%lf",&u,&v,&w,&xp);
xp=-log(1-xp);
add_edge(u,v,w-1,xp);
add_edge(u,v,1,0.0);
}
//cout<<"&&&&\n";
double ans=-mincostflow();
printf("%.2f\n",1-exp(ans));
}
return 0;
}
hdu-5988 Coding Contest(费用流)的更多相关文章
- HDU 5988 Coding Contest(费用流+浮点数)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5988 题目大意: 给定n个点,m条有向边,每个点是一个吃饭的地方,每个人一盒饭.每个点有S个人,有B盒 ...
- HDU 5988.Coding Contest 最小费用最大流
Coding Contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)To ...
- HDU 5988 Coding Contest(浮点数费用流)
http://acm.split.hdu.edu.cn/showproblem.php?pid=5988 题意:在acm比赛的时候有多个桌子,桌子与桌子之间都有线路相连,每个桌子上会有一些人和一些食物 ...
- HDU 5988 Coding Contest(最小费用最大流变形)
Problem DescriptionA coding contest will be held in this university, in a huge playground. The whole ...
- Coding Contest(费用流变形题,double)
Coding Contest http://acm.hdu.edu.cn/showproblem.php?pid=5988 Time Limit: 2000/1000 MS (Java/Others) ...
- 2016青岛区域赛.Coding Contest(费用流 + 概率计算转换为加法计算)
Coding Contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)To ...
- HDU5988 - 2016icpc青岛 - G - Coding Contest 费用流(利用对数化乘为加
HDU5988 题意: 有n个区域,每个区域有s个人,b份饭.现在告诉你每个区域间的有向路径,每条路有容量和损坏路径的概率.问如何走可以使得路径不被破坏的概率最小.第一个人走某条道路是百分百不会损坏道 ...
- HDU 5988 Coding Contest 最小费用流 cost->double
Problem Description A coding contest will be held in this university, in a huge playground. The whol ...
- HDU5988 Coding Contest(费用流)
2016青岛现场赛的一题,由于第一次走过不会产生影响,需要拆点,不过比赛时没想到,此外还有许多细节要注意,如要加eps,时间卡得较紧要注意细节优化等 #include <iostream> ...
随机推荐
- 5G
前世 1G 2G 3G 4G 今生 5G 推荐书籍: Gold Smith -<无线通信> David Tse -<无线通信基础> 参考:
- EntityFrame Work 6 Code First 配置字段为varchar 类型
EntityFrame Work 6 配置字符串属性是否支持Unicode 内容 默认情况下,字符串为Unicode(SQLServer 中的nvarchar).您可以使用IsUnicode 方法指定 ...
- sql语句查询
1. sql语句查询某位数字或者某几位数字开头的数据,字段类型为数字类: %’: 2. sql搜索以4开头和含有李字的数据: select * from wlzbpre_user where real ...
- Difficult Melody(映射)
题目链接 http://vjudge.net/contest/137242#problem/D Description You're addicted to a little game called ...
- php实现设计模式之 原型模式
<?php /* * 原型模式:通过复制已经存在的对象来创建新对象. * 通过原型实例指定创建对象的种类,并且通过copy这些原型创建信的对象 * 是创建型模式 * 有的时候创建一个对象有很多步 ...
- 大公司c#&.net转型java的原因有哪些?
历来就听说有编程语言“鄙视链”的说法,而如今月经贴上的那些事儿,还真让我给遇到了. 以下内容来自知乎,纯属扯淡,易引发口水战,看完勿人身攻击. 目的给盲目的公司决策者.开发人员科普下,有个客观清醒的认 ...
- Hibernate总结(一)
Hibernate为了提高性能,提供了缓存与快照机制. 它的缓存分为一级缓存与二级缓存. Hibernate一级缓存:当一个事务中执行一次Sql语句时,就将返回的结果存储在Session中的Map集合 ...
- 浅谈tornado项目应用设计
一.预备知识 最近开始尝试做一些tornado商城项目,在开始之前需要引入一些项目设计知识,如接口,抽象方法抽象类,组合,程序设计原则等,个人理解项目的合理设计可增加其灵活性,降低数据之间的耦合性,提 ...
- 基于WCF MSMQ 的企业应用解决方案
最近研究了一下基于MSMQ的WCF应用,从书上.网上查了很多资料,但始终没能彻底理解WCF-MSMQ的工作原理,也没能得到一个合理的应用解决方案.索性还是自己做个实验,探索一下吧.经过反复试验,颇有收 ...
- 简要分析webpack打包后代码
开门见山 1.打包单一模块 webpack.config.js module.exports = { entry:"./chunk1.js", output: { path: __ ...