Rain on your Parade

https://vjudge.net/contest/238849#problem/F

题目的意思就是再有t秒就要下雨了，宾客去拿伞，问最多有多少宾客可以拿到伞

二分匹配，用匈牙利算法一定会超时，就用Hk算法

#include
#include
#include
#include
#include
#include
const int inf=0x3f3f3f3f;
using namespace std;
int g[3010][3010];
int dx[3010];
int dy[3010];
int linex[3010];
int liney[3010];
int vis[3010];
int dis;
int t,n,m;
struct node1
{
    int x;
    int y;
    int v;
}people[3010];
struct node2
{
    int x;
    int y;
}umbrella[3010];
int abs(int a,int b)
{
    if(a>b)
        return a-b;
    else
        return b-a;
}
int distance(node1 a,node2 b)
{
    return abs(a.x-b.x)+abs(a.y-b.y);
}
bool searchpath()
{
    queueq;
    dis=inf;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));
    for(int i=1;i<=n;i++)
    {
        if(linex[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=1;i<=m;i++)
        {
            if(g[u][i]&&dy[i]==-1)
            {
                dy[i]=dx[u]+1;
                if(liney[i]==-1)dis=dy[i];
                else
                {
                    dx[liney[i]]=dy[i]+1;
                    q.push(liney[i]);
                }
            }
        }
    }
    return dis!=inf;
}
int find1(int v)
{
    for(int i=1;i<=m;i++)
    {
        if(!vis[i]&&g[v][i]&&dy[i]==dx[v]+1)
        {
            vis[i]=1;
            if(liney[i]!=-1&&dy[i]==dis)continue;
            if(liney[i]==-1||find1(liney[i]))
            {
                liney[i]=v;
                linex[v]=i;
                return 1;
            }
        }
    }
    return 0;
}
int maxmatch()
{
    int ans=0;
    memset(linex,-1,sizeof(linex));
    memset(liney,-1,sizeof(liney));
    while(searchpath())
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            if(linex[i]==-1)
            {
                ans+=find1(i);
            }
        }
    }

    return ans;
}
int main()
{
    int T;
    int r=0;
    scanf("%d",&T);
    while(T--)
    {
        memset(g,0,sizeof(g));
        scanf("%d%d",&t,&n);
        for(int i=1;i<=n;i++)
            scanf("%d%d%d",&people[i].x,&people[i].y,&people[i].v);
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
            scanf("%d%d",&umbrella[i].x,&umbrella[i].y);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(distance(people[i],umbrella[j])/people[i].v<=t)//如果在t时间内能拿到伞就将人伞联系在一起
                    g[i][j]=1;
            }
        }
        printf("Scenario #%d:\n%d\n\n",++r,maxmatch());
    }
}