//640MS	296K
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<map>
#include<stack>
#include<queue>
#include<vector>

#pragma comment(linker, "/STACK:1024000000");
#define M 100007
#define inf 0x3f3f3f3f
#define ll long long
#define mod 1000000009
#define eps (1e-8)
using namespace std;
int ans[1500];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
      memset(ans,0,sizeof(ans));
      int n,num,s_hour,s_minute,e_hour,e_minute,minn=inf,maxx=-1,count=-1;
      scanf("%d",&n);
      for(int i=0;i<n;i++)
      {
          scanf("%d %d:%d%d:%d",&num,&s_hour,&s_minute,&e_hour,&e_minute);
          int l=s_hour*60+s_minute;
          int r=e_hour*60+e_minute;
          if(l<minn)minn=l;
          if(r>maxx)maxx=r;
          for(int j=l;j<r;j++)
            ans[j]+=num;
      }
      for(int i=minn;i<=maxx;i++)
        if(ans[i]>count)count=ans[i];
      printf("%d\n",count);
   }
  return 0;
}

TIANKENG’s travel

2
2 1000
0 0
2000 1000
1000 0
2000 0
0 1000
0 0
2000 1000

2
impossible

推断三点是否一条直线，能够用叉积推断。

//171MS	4320K
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define M 1017
#define inf 0x3f3f3f3f
#define ll long long
#pragma comment(linker, "/STACK:1024000000");
using namespace std;
int t,n;
long long l;
int s,e,num=0;
int dist[M] ,vis[M] ,head[1001005];
struct sa
{
    int id;
    long long x,y;
}node[M] ;
struct E
{
    int v,next,cap;
}Edge[1001005];
void addEdge(int u,int v,int c)
{
    Edge[num].v=v;Edge[num].cap=c;
    Edge[num].next=head[u];
    head[u]=num++;
}
int cmp(sa a,sa b)
{
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
long long judge(long long  a,long long b)
{
    long long x=(node[a] .x-node[b] .x)*(node[a] .x-node[b] .x);
    long long y=(node[a] .y-node[b] .y)*(node[a] .y-node[b] .y);
    long long len=l*l;
    return x+y;
}
void SPFA()                   //  源点为0，汇点为sink。
{
    queue<int>q;
    for(int i=0;i<n;i++)dist[i]=inf;
    q.push(s);
    dist[s] = 0;
    vis[s] =1;
    while(!q.empty())        //  这里最好用队列。有广搜的意思，堆栈像深搜。

    {
        int u =q.front();
        q.pop();vis[u]=0;
        for(int i=head[u]; i!=-1;i=Edge[i].next)
        {
            int v=Edge[i].v;
            if(dist[v]>dist[u]+1)
            {
                dist[v]=dist[u]+1;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=true;
                }
            }
        }
    }
}
bool cha(int i,int j,int k)
{
    int x1=node[i] .x-node[j] .x;
    int x2=node[j] .x-node[k] .x;
    int y1=node[i] .y-node[j] .y;
    int y2=node[j] .y-node[k] .y;
    if(x1*y2-x2*y1)return true;
    return false;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(head,-1,sizeof(head));
        num=0;
        scanf("%d%I64d",&n,&l);
        n+=2;
       for(int i=0;i<n;i++)
        {
            scanf("%d%d",&node[i] .x,&node[i] .y);
            node[i] .id=i;
        }
        sort(node,node+n,cmp);
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                if(judge(i,j)<=l*l)
                {
                    int flag=1;
                    for(int k=i+1;k<j;k++)
                        if(!cha(i,j,k)){flag=0;break;}
                    if(flag){addEdge(i,j,1);addEdge(j,i,1);}
                }
        for(int i=0;i<n;i++)
        {
            if(node[i].id==0)s=i;
            if(node[i].id==1)e=i;
        }
        SPFA();
        if(dist[e] ==inf)printf("impossible\n");
        else printf("%d\n",dist[e] -1);
    }
    return 0;
}