2018.07.06 POJ2536 Gopher II(二分图匹配)
Gopher II
Time Limit: 2000MS Memory Limit: 65536K
Description
The gopher family, having averted the canine threat, must face a new predator.
The are n gophers and m gopher holes, each at distinct (x, y) coordinates. A hawk arrives and if a gopher does not reach a hole in s seconds it is vulnerable to being eaten. A hole can save at most one gopher. All the gophers run at the same velocity v. The gopher family needs an escape strategy that minimizes the number of vulnerable gophers.
Input
The input contains several cases. The first line of each case contains four positive integers less than 100: n, m, s, and v. The next n lines give the coordinates of the gophers; the following m lines give the coordinates of the gopher holes. All distances are in metres; all times are in seconds; all velocities are in metres per second.
Output
Output consists of a single line for each case, giving the number of vulnerable gophers.
Sample Input
2 2 5 10
1.0 1.0
2.0 2.0
100.0 100.0
20.0 20.0
Sample Output
1
Source
Waterloo local 2001.01.27
一道二分图匹配的板子题,感觉dinic" role="presentation" style="position: relative;">dinicdinic算法快的飞起,于是我写了个dinic" role="presentation" style="position: relative;">dinicdinic求二分图最大匹配,其他没什么,就是建图的时候要记住判定地鼠到洞的距离是否合法就行了,还有就是多组数据记得要重置数组和cnt" role="presentation" style="position: relative;">cntcnt
代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#define N 300
#define M 100000
using namespace std;
struct pot{double x,y;}p[N];
int n,m,T,V,s,t,cnt,first[N],d[N];
struct edge{int v,next,c;}e[M];
inline void add(int u,int v,int c){
e[++cnt].v=v;
e[cnt].c=c;
e[cnt].next=first[u];
first[u]=cnt;
}
inline bool bfs(){
queue<int>q;
memset(d,-1,sizeof(d));
d[s]=0;
q.push(s);
while(!q.empty()){
int x=q.front();
q.pop();
for(int i=first[x];i!=-1;i=e[i].next){
int v=e[i].v;
if(d[v]!=-1||e[i].c<=0)continue;
d[v]=d[x]+1;
if(v==t)return true;
q.push(v);
}
}
return false;
}
inline int dfs(int p,int f){
if(p==t||!f)return f;
int flow=f;
for(int i=first[p];i!=-1;i=e[i].next){
int v=e[i].v;
if(d[v]==d[p]+1&&e[i].c>0&&flow){
int tmp=dfs(v,min(flow,e[i].c));
if(!tmp)d[v]=-1;
e[i].c-=tmp;
e[i^1].c+=tmp;
flow-=tmp;
}
}
return f-flow;
}
inline double dis(pot a,pot b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
int main(){
while(~scanf("%d%d%d%d",&n,&m,&T,&V)){
s=0,t=n+m+1;
int len=T*V;
cnt=-1;
memset(first,-1,sizeof(first));
for(int i=1;i<=n+m;++i)scanf("%lf%lf",&p[i].x,&p[i].y);
for(int i=1;i<=n;++i)add(s,i,1),add(i,s,0);
for(int i=n+1;i<=n+m;++i)add(i,t,1),add(t,i,0);
for(int i=1;i<=n;++i)
for(int j=n+1;j<=n+m;++j)
if(dis(p[i],p[j])<=len*1.0)add(i,j,1),add(j,i,0);
int ans=0;
while(bfs())ans+=dfs(s,0x3f3f3f3f);
printf("%d\n",n-ans);
}
return 0;
}
2018.07.06 POJ2536 Gopher II(二分图匹配)的更多相关文章
- POJ2536 Gopher II(二分图最大匹配)
Gopher II Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9005 Accepted: 3724 Descrip ...
- poj 2536 Gopher II (二分匹配)
Gopher II Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6345 Accepted: 2599 Descrip ...
- HDU-3081-Marriage Match II 二分图匹配+并查集 OR 二分+最大流
二分+最大流: 1 //题目大意:有编号为1~n的女生和1~n的男生配对 2 // 3 //首先输入m组,a,b表示编号为a的女生没有和编号为b的男生吵过架 4 // 5 //然后输入f组,c,d表示 ...
- 2018.06.27Going Home(二分图匹配)
Going Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24716 Accepted: 12383 Descript ...
- POJ2536 Gopher II【二分图最大匹配】
题目链接: http://poj.org/problem? id=2536 题目大意: 有N仅仅鼹鼠和M个洞穴,假设鼹鼠在S秒内不可以跑到洞穴,就会被老鹰捉住吃掉. 鼹鼠跑的速度为V米/秒. 已知一个 ...
- 2018.07.06 POJ1698 Alice's Chance(最大流)
Alice's Chance Time Limit: 1000MS Memory Limit: 10000K Description Alice, a charming girl, have been ...
- 2018.07.06 POJ1556 The Doors(最短路)
The Doors Time Limit: 1000MS Memory Limit: 10000K Description You are to find the length of the shor ...
- EZ 2018 07 06 NOIP模拟赛
又是慈溪那边给的题目,这次终于没有像上次那样尴尬了, T1拿到了较高的暴力分,T2没写炸,然后T3写了一个优雅的暴力就203pts,Rank3了. 听说其它学校的分数普遍100+,那我们学校还不是强到 ...
- 2018.07.06 BZOJ 1588: HNOI2002营业额统计(非旋treap)
1588: [HNOI2002]营业额统计 Time Limit: 5 Sec Memory Limit: 162 MB Description 营业额统计 Tiger最近被公司升任为营业部经理,他上 ...
随机推荐
- as3 运算与检查String 是否能够正确转换成数 值
如果忘了对一个Number 型变量初始化,那么这个变量参与的任何数学运算的结果都是NaN:如果最终结果赋值给有声明类型的变量,那么为该变量的默认值(仅限uint ,int). var a:Number ...
- eclipse xDoclet错误
转载: http://blog.csdn.net/lovelongjun/article/details/53485773 Missing library: xdoclet-1.2.1.jar. Se ...
- ABAP-长文本处理
- apiCloud事件发送与监听
apiCloud事件发送与监听 1.sendEvent 将任意一个自定义事件广播出去,该事件可在任意页面通过 addEventListener 监听收到. sendEvent({params}) 2. ...
- kernel TCP time wait bucket table overflow
# 故障描述 有一个需求是实时分析API接口访问日志,提取token去数据库查询对应的uid,然后收集一些指标存入到hbase中. 当程序执行一会后会被系统杀死 Killed ! # 故障排查 .CP ...
- Kotlin语言学习笔记(7)
反射 // 反射 val c = MyClass::class val c2 = MyClass::class.java // 获取KClass的引用 val widget: Widget = ... ...
- 初始化centoS 相关
install aspnetcoremodule for iis https://docs.microsoft.com/en-us/aspnet/core/publishing/iis?tabs=as ...
- sessionStorage和localStorage
html5中的Web Storage包括了两种存储方式:sessionStorage和localStorage. sessionStorage用于本地存储一个会话(session)中的数据,这些数据只 ...
- 必备 .NET - C# 脚本
作者:Mark Michaelis | 2016 年 1 月 Link: https://msdn.microsoft.com/zh-cn/magazine/mt614271.aspx 随着 Visu ...
- Python3 ssl模块不可用的问题
编译安装完Python3之后,使用pip来安装python库,发现了如下报错: $ pip install numpy pip is configured with locations that re ...