Rain on your Parade

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 655350/165535 K (Java/Others)
Total Submission(s): 3033 Accepted Submission(s): 952


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 題意：第一行案例數，每個案例第一行 代表還有多少單位時間開始下雨，然後是 N個訪客，接下來N行是 每個訪客的位置（一維坐標平面內）和他的移動速度，接下來M行 代表雨傘數目，接下來M行表示各個雨傘的位置，問在下雨前 最多有多少人能夠拿到雨傘（兩個人不能共用一把傘）。
解析：以在給定的時間內 人能到達傘的位置 建邊，二分最大匹配，如果直接用經典匈牙利算法一定分超時的，可用Hopcroft-Karp算法（O（sqrt(n)*edgnum））. 546MS AC。

#include
#include
#include
#include
using namespace std;
const int N = 3005;
const int INF=1<<28;

int head[N],to[N*N],next1[N*N],tot;  //存圖
int dx[N],dy[N],dis;                 //左邊部分距離，右邊部分距離，記錄右邊部分沒有被匹配過的點的最大距離
int machx[N],nx,machy[N];            //左邊部分點匹配右邊點，左邊部分的點個數，右邊部分點匹配左邊的點
bool vist[N];

void addEdg(int u,int v){
    to[tot]=v; next1[tot]=head[u]; head[u]=tot++;
}
bool searchpath(){//找有沒有增廣路
    queueq;
    dis=INF;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));

    for(int i=1; i<=nx; i++)
        if(machx[i]==-1)
        q.push(i),dx[i]=0;

    while(!q.empty()){
        int u=q.front(); q.pop();
        if(dx[u]>dis)
            break;
        for(int i=head[u]; i!=-1; i=next1[i]){
            int v=to[i];
            if(dy[v]==-1){
                dy[v]=dx[u]+1;
                if(machy[v]==-1)
                    dis=dy[v];
                else{
                    dx[machy[v]]=dy[v]+1;
                    q.push(machy[v]);
                }
            }
        }
    }
    return dis!=INF;
}
bool findroad(int u){
    for(int i=head[u]; i!=-1; i=next1[i]){
        int v=to[i];
        if(!vist[v]&&dy[v]==dx[u]+1){
            vist[v]=1;
            if(machy[v]!=-1&&dy[v]==dis)
                continue;
            if(machy[v]==-1||findroad(machy[v])){
                machy[v]=u; machx[u]=v;  return true;
            }
        }
    }
    return false;
}
int MaxMatch(){
    int ans=0;
    memset(machx,-1,sizeof(machx));
    memset(machy,-1,sizeof(machy));
    while( searchpath() ){
        memset(vist,0,sizeof(vist));
        for(int i=1; i<=nx; i++)
            if(machx[i]==-1)
            ans+=findroad(i);
    }
    return ans;
}
//-------------上面部分的代碼為模板---------------------

struct node{
    int x,y;
    double dis;
}man[N],umb[N];

double countDis(int u,int v){
    return sqrt((man[u].x-umb[v].x)*(man[u].x-umb[v].x)*1.0+(man[u].y-umb[v].y)*(man[u].y-umb[v].y)*1.0);
}
int main(){
    int T,ny,tim,c=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&tim,&nx);
        for(int i=1;i<=nx;i++){
            scanf("%d%d%lf",&man[i].x,&man[i].y,&man[i].dis);
            man[i].dis*=tim;
        }
        scanf("%d",&ny);
        for(int i=1;i<=ny;i++)
            scanf("%d%d",&umb[i].x,&umb[i].y);

//---------------------建圖---------------------
        tot=0;
        memset(head,-1,sizeof(head));
        for(int u=1;u<=nx;u++)
            for(int v=1;v<=ny;v++)
            if(man[u].dis>=countDis(u,v))
                addEdg(u,v);
//----------------------------------------------
        int ans=MaxMatch();
        printf("Scenario #%d:\n%d\n\n",++c,ans);
    }
    return 0;
}