Rain on your Parade

Rain on your Parade

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)
Total Submission(s): 2113    Accepted Submission(s): 647

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.

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

一看就是求二分图的最大匹配,不过点太多了匈牙利不行,要用Hopcroft_Karp.别人的代码没看懂,不过幸好接口用起来挺方便.贴上大牛的代码:

#include<cstring>   //HK算法。。
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>   

using namespace std;
const int maxn=;
int n,m;
struct Pos{
    int x,y;
}man[maxn],umbr[maxn];
vector<int> child[maxn];
int speed[maxn];//人的速度
int cx[maxn],cy[maxn];//集合X Y的匹配值
int distx[maxn],disty[maxn];//记录距离   

bool bfs(){
    bool flag=false;
    memset(distx,,sizeof(distx));//距离初始为0
    memset(disty,,sizeof(disty));  

    queue<int> que;
    for(int i=;i<=n;i++)//把X集合中    所有未匹配的加入队列
        if(cx[i]==-)
            que.push(i);
    while(!que.empty())//集合非空
    {
        int x=que.front();
        que.pop();
        for(int i =  ; i < child[ x ].size() ; i++ )//所有以i为起点的边
        {
            int y = child[ x ][ i ];
            if( !disty[ y ] )//该边未被使用
            {
                    disty[ y ]=distx[ x ] + ;//距离+1
                    if(cy[ y ] == -) flag=true;//该点未被使用可以增加匹配
                    else
                    {
                        distx[ cy[ y ] ]= disty[ y ] + ;
                        que.push( cy[ y ] );
                    }
                }
            }
        }
    return flag;
}
bool dfs(int x){//寻找增广路
    for(int i=;i<child[x].size();i++)//枚举所有以i为起点的边
    {
        int y=child[x][i];
        if(disty[ y ] == distx[ x ]+)//距离刚好为1表明相连
        {
            disty[ y ] = ;
            if(cy[ y ] == - || dfs( cy[ y ] ))//找到一条增广路
            {
                cx[ x ] = y ; cy[ y ] = x;//保存匹配值
                return true;
            }
        }
    }
    return false;
}
int Hopcroft_Karp(){
    memset(cx,-,sizeof(cx));
    memset(cy,-,sizeof(cy));
    int ans=;
    while(bfs())//如果距离更新成功，X集合中每一个点找一次增广路
    {
        for(int i=;i<=n;i++)
            if(cx[i]==- && dfs(i))
                ans++;
    }
    return ans;
}  

double dis(int i,int j){
    return sqrt( (double)(man[i].x-umbr[j].x)*(man[i].x-umbr[j].x)+ (man[i].y-umbr[j].y)*(man[i].y-umbr[j].y) );
}
int main(){
    int t,cas=,time;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&time,&n);
        for(int i=;i<=n;i++)
            scanf("%d%d%d",&man[i].x,&man[i].y,&speed[i]);  

        scanf("%d",&m);
        for(int i=;i<=m;i++)
            scanf("%d%d",&umbr[i].x,&umbr[i].y);  

        for(int i=;i<=n;i++)
            child[i].clear();  

        for(int i=;i<=n;i++)//能够在下雨前拿到伞的建一条边
            for(int j=;j<=m;j++)
                if(speed[i]*time >= dis(i,j))
                    child[i].push_back(j);  

        printf("Scenario #%d:\n",++cas);
        printf("%d\n",Hopcroft_Karp());
        printf("\n");  

    }
}