hdu 2389(二分图hk算法模板)

Rain on your Parade

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)
Total Submission(s): 3675    Accepted Submission(s): 1187

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 <= 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

 

题意:n个人匹配m把伞,问最多能有多少匹配?

题解:HK算法减少时间.

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
const int N = ;
const int INF = ;
struct Node
{
    int x,y,v;
} p[N],ub[N];
int graph[N][N];
int n,m,dist;
bool vis[N];
int cx[N],cy[N],dx[N],dy[N];
int dis(Node a,Node b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
bool searchpath()
{
    queue<int> Q;
    dist=INF;
    memset(dx,-,sizeof(dx));
    memset(dy,-,sizeof(dy));
    for(int i=;i<=n;i++)
    {
        if(cx[i]==-)
        {
            Q.push(i);
            dx[i]=;
        }
    }
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        if(dx[u]>dist) break;
        for(int v=;v<=n;v++)
        {
            if(graph[u][v]&&dy[v]==-)
            {
                dy[v]=dx[u]+;
                if(cy[v]==-) dist=dy[v];
                else
                {
                    dx[cy[v]]=dy[v]+;
                    Q.push(cy[v]);
                }
            }
        }
    }
    return dist!=INF;
}
int findpath(int u)
{
    for(int v=;v<=m;v++)
    {
        if(!vis[v]&&graph[u][v]&&dy[v]==dx[u]+)
        {
            vis[v]=;
            if(cy[v]!=-&&dy[v]==dist)
            {
                continue;
            }
            if(cy[v]==-||findpath(cy[v]))
            {
                cy[v]=u;cx[u]=v;
                return ;
            }
        }
    }
    return ;
}
void MaxMatch()
{
    int res=;
    memset(cx,-,sizeof(cx));
    memset(cy,-,sizeof(cy));
    while(searchpath())
    {
        memset(vis,,sizeof(vis));
        for(int i=;i<=n;i++)
        {
            if(cx[i]==-)
            {
                res+=findpath(i);
            }
        }
    }
    printf("%d\n\n",res);;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int t = ;
    while(tcase--)
    {
        memset(graph,,sizeof(graph));
        int time;
        scanf("%d",&time);
        scanf("%d",&n);
        for(int i=; i<=n; i++)
        {
            int v;
            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].v);
        }
        scanf("%d",&m);
        for(int i=; i<=m; i++)
        {
            scanf("%d%d",&ub[i].x,&ub[i].y);
        }
        for(int i=; i<=n; i++)
        {
            for(int j=; j<=m; j++)
            {
                if(dis(p[i],ub[j])<=p[i].v*p[i].v*time*time)
                {
                    graph[i][j] = ;
                }
            }
        }
        printf("Scenario #%d:\n",t++);
        MaxMatch();
    }
    return ;
}