二分图------》Hopcroft-Karp算法 hdu2389

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

 

#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;  
}