hdu 2389 Rain on your Parade

Problem 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 <= si
 <= 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.
// 这题是很裸的二分匹配了 主要是HK二分匹配算法的模板 
// 表示不清楚为什么会比直接dfs快蛮多、不会证明 呵呵
#include <iostream>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N=3010;
struct cor
{
    int x,y;
}X[N],Y[N];
int speed[N];
bool visit[N];
int cx[N],cy[N];
int dx[N],dy[N];

vector <int> G[N];
double dis(int i,int j)
{
   return sqrt(double( (X[i].x-Y[j].x)*(X[i].x-Y[j].x) + (X[i].y-Y[j].y)*(X[i].y-Y[j].y) ) );
}
int m,n;
bool search()
{
    memset(dx,0,sizeof(dx));
    memset(dy,0,sizeof(dy));
    queue <int > Q;
    int i,u,v;
    bool flag=0;

    for(i=1;i<=m;i++)
     if(!cx[i]) Q.push(i);

    while(!Q.empty())
    {
        u=Q.front();Q.pop();
        for(i=0;i<G[u].size();i++)
        {
            v=G[u][i];
            if(!dy[v])
            {
                dy[v]=dx[u]+1;
                if(!cy[v])
                   flag=true;
                else
                {
                   dx[cy[v]]=dy[v]+1;
                   Q.push(cy[v]);
                }
            }
        }
    }
    return flag;
}
bool dfs(int u)
{
    int i,v;
    for(i=0;i<G[u].size();i++)
     {
         v=G[u][i];
         if(!visit[v]&&dy[v]==dx[u]+1)
         {
              visit[v]=true;
              if(!cy[v]||dfs(cy[v]))
              {
                  cy[v]=u;
                  cx[u]=v;
                  return true;
              }
         }
     }
     return false;
}

int Maxmatch()
{
    memset(cx,0,sizeof(cx));
    memset(cy,0,sizeof(cy));
    int ans=0;
    while(search())
    {
        memset(visit,0,sizeof(visit));
        for(int i=1;i<=m;i++)
         if(!cx[i]&&dfs(i))
           ans++;
    }
   return ans;
}
int main()
{
    int t,cnt=1;
    scanf("%d",&t);
    while(t--)
    {
       int tm;
       int i,j;
       scanf("%d",&tm);
       scanf("%d",&m);
       for(i=1;i<=m;i++)
        scanf("%d %d %d",&X[i].x,&X[i].y,&speed[i]);
       scanf("%d",&n);
       for(i=1;i<=n;i++)
        scanf("%d %d",&Y[i].x,&Y[i].y);
       for(i=1;i<=m;i++)
        {
            G[i].clear();
            for(j=1;j<=n;j++)
             if(speed[i]*tm>=dis(i,j))
               G[i].push_back(j);
        }
       printf("Scenario #%d:\n",cnt++);
       printf("%d\n\n",Maxmatch());
    }

    return 0;
}