二分图------》Hopcroft-Karp算法 hdu2389
2016-03-07 23:11
323 查看
Hopcroft-Karp算法
该算法由John.E.Hopcroft和Richard M.Karp于1973提出,故称Hopcroft-Karp算法。
原理
为了降低时间复杂度,可以在增广匹配集合M时,每次寻找多条增广路径。这样就可以进一步降低时间复杂度,可以证明,算法的时间复杂度可以到达O(n^0.5*m),虽然优化不了多少,但在实际应用时,效果还是很明显的。
基本算法
该算法主要是对匈牙利算法的优化,在寻找增广路径的时候同时寻找多条不相交的增广路径,形成极大增广路径集,然后对极大增广路径集进行增广。在寻找
增广路径集的每个阶段,找到的增广路径集都具有相同的长度,且随着算法的进行,增广路径的长度不断的扩大。可以证明,最多增广n^0.5次就可以得到最大
匹配。
算法流程
(1)从G=(X,Y;E)中取一个初始匹配。
(2)若X中的所有顶点都被M匹配,则表明M为一个完美匹配,返回;否则,以所有未匹配顶点为源点进行一次BFS,标记各个点到源点的距离。
(3)在满足dis[v] = dis[u] + 1的边集<v,u>中,从X中找到一个未被M匹配的顶点x0,记S = {x0},T = ¢。
(4)若N(S) = T,则表明当前已经无法得到更大匹配,返回;否则取一y0∈N(S) - 。
(5)若y0已经被M匹配则转步骤(6),否则做一条x0->y0的M-增广路径P(x0,y0),取M = M△P(x0,y0)。
(6)由于y已经被M匹配,所以M中存在一条边(y0,z0)去S = S∪ {z0},T = T∪{y0},转步骤(2)。
算法具体时间与分析
在寻找增广路径中可以对X中的每个未匹配的顶点进行BFS,BFS时对每个顶点维护一个距离编号dx[nx],dy[ny],如果某个Y中的节点为
未匹配点,则找到一条增广路径。BFS结束后找到了增广路径集。然后利用DFS与匈牙利算法类似的方法对每条增广路进行增广,这样就可以找到最大匹配。
F - Rain on your Parade
Time Limit:3000MS Memory Limit:165535KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2389
Appoint description:
Description
You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It’s a warm, sunny evening, and a soothing wind sends fresh, salty air from the sea. The evening is progressing just as you had imagined. It could be the perfect end of a beautiful day.
But nothing ever is perfect. One of your guests works in
weather forecasting. He suddenly yells, “I know that breeze! It means
its going to rain heavily in just a few minutes!” Your guests all wear
their best dresses and really would not like to get wet, hence they
stand terrified when hearing the bad news.
You have prepared a few umbrellas which can protect a few of
your guests. The umbrellas are small, and since your guests are all
slightly snobbish, no guest will share an umbrella with other guests.
The umbrellas are spread across your (gigantic) garden, just like your
guests. To complicate matters even more, some of your guests can’t run
as fast as the others.
Can you help your guests so that as many as possible find an umbrella before it starts to pour?
Given the positions and speeds of all your guests, the
positions of the umbrellas, and the time until it starts to rain, find
out how many of your guests can at most reach an umbrella. Two guests do
not want to share an umbrella, however.
Input
The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line containing the time t in
minutes until it will start to rain (1 <=t <= 5). The next line
contains the number of guests m (1 <= m <= 3000), followed by m
lines containing x- and y-coordinates as well as the speed si in units
per minute (1 <= s i <= 3000) of the guest as integers,
separated by spaces. After the guests, a single line contains n (1
<= n <= 3000), the number of umbrellas, followed by n lines
containing the integer coordinates of each umbrella, separated by a
space.
The absolute value of all coordinates is less than 10000.
Output
For each test case, write a line containing “Scenario #i:”, where i is
the number of the test case starting at 1. Then, write a single line
that contains the number of guests that can at most reach an umbrella
before it starts to rain. Terminate every test case with a blank line.
Sample Input
2
1
2
1 0 3
3 0 3
2
4 0
6 0
1
2
1 1 2
3 3 2
2
2 2
4 4
Sample Output
Scenario #1:
2
Scenario #2:
2
该算法由John.E.Hopcroft和Richard M.Karp于1973提出,故称Hopcroft-Karp算法。
原理
为了降低时间复杂度,可以在增广匹配集合M时,每次寻找多条增广路径。这样就可以进一步降低时间复杂度,可以证明,算法的时间复杂度可以到达O(n^0.5*m),虽然优化不了多少,但在实际应用时,效果还是很明显的。
基本算法
该算法主要是对匈牙利算法的优化,在寻找增广路径的时候同时寻找多条不相交的增广路径,形成极大增广路径集,然后对极大增广路径集进行增广。在寻找
增广路径集的每个阶段,找到的增广路径集都具有相同的长度,且随着算法的进行,增广路径的长度不断的扩大。可以证明,最多增广n^0.5次就可以得到最大
匹配。
算法流程
(1)从G=(X,Y;E)中取一个初始匹配。
(2)若X中的所有顶点都被M匹配,则表明M为一个完美匹配,返回;否则,以所有未匹配顶点为源点进行一次BFS,标记各个点到源点的距离。
(3)在满足dis[v] = dis[u] + 1的边集<v,u>中,从X中找到一个未被M匹配的顶点x0,记S = {x0},T = ¢。
(4)若N(S) = T,则表明当前已经无法得到更大匹配,返回;否则取一y0∈N(S) - 。
(5)若y0已经被M匹配则转步骤(6),否则做一条x0->y0的M-增广路径P(x0,y0),取M = M△P(x0,y0)。
(6)由于y已经被M匹配,所以M中存在一条边(y0,z0)去S = S∪ {z0},T = T∪{y0},转步骤(2)。
算法具体时间与分析
在寻找增广路径中可以对X中的每个未匹配的顶点进行BFS,BFS时对每个顶点维护一个距离编号dx[nx],dy[ny],如果某个Y中的节点为
未匹配点,则找到一条增广路径。BFS结束后找到了增广路径集。然后利用DFS与匈牙利算法类似的方法对每条增广路进行增广,这样就可以找到最大匹配。
F - Rain on your Parade
Time Limit:3000MS Memory Limit:165535KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2389
Appoint description:
Description
You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It’s a warm, sunny evening, and a soothing wind sends fresh, salty air from the sea. The evening is progressing just as you had imagined. It could be the perfect end of a beautiful day.
But nothing ever is perfect. One of your guests works in
weather forecasting. He suddenly yells, “I know that breeze! It means
its going to rain heavily in just a few minutes!” Your guests all wear
their best dresses and really would not like to get wet, hence they
stand terrified when hearing the bad news.
You have prepared a few umbrellas which can protect a few of
your guests. The umbrellas are small, and since your guests are all
slightly snobbish, no guest will share an umbrella with other guests.
The umbrellas are spread across your (gigantic) garden, just like your
guests. To complicate matters even more, some of your guests can’t run
as fast as the others.
Can you help your guests so that as many as possible find an umbrella before it starts to pour?
Given the positions and speeds of all your guests, the
positions of the umbrellas, and the time until it starts to rain, find
out how many of your guests can at most reach an umbrella. Two guests do
not want to share an umbrella, however.
Input
The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line containing the time t in
minutes until it will start to rain (1 <=t <= 5). The next line
contains the number of guests m (1 <= m <= 3000), followed by m
lines containing x- and y-coordinates as well as the speed si in units
per minute (1 <= s i <= 3000) of the guest as integers,
separated by spaces. After the guests, a single line contains n (1
<= n <= 3000), the number of umbrellas, followed by n lines
containing the integer coordinates of each umbrella, separated by a
space.
The absolute value of all coordinates is less than 10000.
Output
For each test case, write a line containing “Scenario #i:”, where i is
the number of the test case starting at 1. Then, write a single line
that contains the number of guests that can at most reach an umbrella
before it starts to rain. Terminate every test case with a blank line.
Sample Input
2
1
2
1 0 3
3 0 3
2
4 0
6 0
1
2
1 1 2
3 3 2
2
2 2
4 4
Sample Output
Scenario #1:
2
Scenario #2:
2
#include<stdio.h> #include<queue> #include<iostream> #include<string.h> #include<math.h> using namespace std; #define eps 1e-6 const int MAXN=3005; const int INF=1<<28; int g[MAXN][MAXN],Mx[MAXN],My[MAXN],Nx,Ny; int dx[MAXN],dy[MAXN],dis; bool vst[MAXN]; struct Node1 { int x,y,s; }guests[MAXN]; struct Node2 { int x,y; }um[MAXN]; double distance(Node1 a,Node2 b) { double x=a.x-b.x; double y=a.y-b.y; return sqrt(x*x+y*y); } bool searchP() { queue<int>Q; dis=INF; memset(dx,-1,sizeof(dx)); memset(dy,-1,sizeof(dy)); for(int i=0;i<Nx;i++) if(Mx[i]==-1) { Q.push(i); dx[i]=0; } while(!Q.empty()) { int u=Q.front(); Q.pop(); if(dx[u]>dis) break; for(int v=0;v<Ny;v++) if(g[u][v]&&dy[v]==-1) { dy[v]=dx[u]+1; if(My[v]==-1) dis=dy[v]; else { dx[My[v]]=dy[v]+1; Q.push(My[v]); } } } return dis!=INF; } bool DFS(int u) { for(int v=0;v<Ny;v++) if(!vst[v]&&g[u][v]&&dy[v]==dx[u]+1) { vst[v]=1; if(My[v]!=-1&&dy[v]==dis) continue; if(My[v]==-1||DFS(My[v])) { My[v]=u; Mx[u]=v; return 1; } } return 0; } int MaxMatch() { int res=0; memset(Mx,-1,sizeof(Mx)); memset(My,-1,sizeof(My)); while(searchP()) { memset(vst,0,sizeof(vst)); for(int i=0;i<Nx;i++) if(Mx[i]==-1&&DFS(i)) res++; } return res; } int main(){ int n,m,t,i,j; int T,iCase=0; scanf("%d",&T); while(T--) { iCase++; scanf("%d",&t); scanf("%d",&m); for(i=0;i<m;i++) scanf("%d%d%d",&guests[i].x,&guests[i].y,&guests[i].s); scanf("%d",&n); for(i=0;i<n;i++) scanf("%d%d",&um[i].x,&um[i].y); Nx=m;Ny=n; memset(g,0,sizeof(g)); for(i=0;i<m;i++) { for(j=0;j<n;j++) { if(distance(guests[i],um[j])/guests[i].s-t<eps) { g[i][j]=1; } } } printf("Scenario #%d:\n%d\n\n",iCase,MaxMatch()); } return 0; }
相关文章推荐
- Linux命令学习笔记之三用户和用户组操作
- 基于tiny4412的Linux内核移植 --- aliases节点解析
- OPP作业
- shell重定向小记
- 在Linux平台上搭建EasyDarwin,编译代码并简单部署
- Educational Codeforces Round 9 E. Thief in a Shop(FFT)
- java -jar shell 启动、停止
- tomcat双击startup.bat闪退的办法
- ecshop显示商品已销售数量
- OPENCV图像轮廓检测
- php网站安全性问题
- OPENCV形态学算法-2
- Linux头文件之stdarg.h
- OpenGL实践之第一个窗口
- OPENCV图像变换-2
- Linux系统平均负载3个数字的含义
- Servlet 工作原理解析
- nginx(五)rewrite
- 高性能Linux服务器 第6章 ext3文件系统反删除利器ext3grep
- linux基础命令