Hdu2389 Rain on your Parade （HK二分图最大匹配）

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 <= 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

 

Source

HDU 2008-10 Public Contest

 

Recommend

lcy

 

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题目的意思是给出那个人的位置和速度，m把伞的位置，问t秒内最多少人拿到伞

思路：拿人和伞进行二分图匹配，数据较大匈牙利炸，用Hopcroft-Karp方法

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;
 
const int MAXN = 3010;//左边节点数量、右边节点数量
const int MAXM = 3010*3010;//边的数量
const int INF = 0x7FFFFFFF;
 
struct Edge
{
    int v;
    int next;
} edge[MAXM];
 
int nx, ny;
int cnt;
int dis;
 
int first[MAXN];
int xlink[MAXN], ylink[MAXN];
/*xlink[i]表示左集合顶点所匹配的右集合顶点序号，ylink[i]表示右集合i顶点匹配到的左集合顶点序号。*/
int dx[MAXN], dy[MAXN];
/*dx[i]表示左集合i顶点的距离编号，dy[i]表示右集合i顶点的距离编号*/
int vis[MAXN]; //寻找增广路的标记数组
 
struct point
{
    int x,y,v;
} a[MAXN],b[MAXN];
 
void init()
{
    cnt = 0;
    memset(first, -1, sizeof(first));
    memset(xlink, -1, sizeof(xlink));
    memset(ylink, -1, sizeof(ylink));
}
 
void read_graph(int u, int v)
{
    edge[cnt].v = v;
    edge[cnt].next = first[u], first[u] = cnt++;
}
 
int bfs()
{
    queue<int> q;
    dis = INF;
    memset(dx, -1, sizeof(dx));
    memset(dy, -1, sizeof(dy));
    for(int i = 0; i < nx; i++)
    {
        if(xlink[i] == -1)
        {
            q.push(i);
            dx[i] = 0;
        }
    }
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(dx[u] > dis) break;
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(dy[v] == -1)
            {
                dy[v] = dx[u] + 1;
                if(ylink[v] == -1) dis = dy[v];
                else
                {
                    dx[ylink[v]] = dy[v]+1;
                    q.push(ylink[v]);
                }
            }
        }
    }
    return dis != INF;
}
 
int find(int u)
{
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(!vis[v] && dy[v] == dx[u]+1)
        {
            vis[v] = 1;
            if(ylink[v] != -1 && dy[v] == dis) continue;
            if(ylink[v] == -1 || find(ylink[v]))
            {
                xlink[u] = v, ylink[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
 
int MaxMatch()
{
    int ans = 0;
    while(bfs())
    {
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < nx; i++) if(xlink[i] == -1)
            {
                ans += find(i);
            }
    }
    return ans;
}
 
int main()
{
    int T,t;
    int q=1;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d",&t);
        scanf("%d",&nx);
        for(int i=0; i<nx; i++)
        {
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].v);
        }
        scanf("%d",&ny);
        for(int i=0; i<ny; i++)
        {
            scanf("%d%d",&b[i].x,&b[i].y);
        }
        for(int i=0; i<nx; i++)
            for(int j=0; j<ny; j++)
            {
                if((a[i].x-b[j].x)*(a[i].x-b[j].x)+(a[i].y-b[j].y)*(a[i].y-b[j].y)<=t*a[i].v*t*a[i].v)
                    read_graph(i,j);
            }
 
        int ans = MaxMatch();
 
        printf("Scenario #%d:\n%d\n\n",q++,ans);
    }
 
    return 0;
}