BestCoder Round #2
TIANKENG’s restaurant
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 600 Accepted Submission(s): 280
Each cases has a positive integer n(1<=n<=10000), which means n groups of customer. Then following n lines, each line there is a positive integer Xi(1<=Xi<=100), referring to the sum of the number of the ith group people, and the arriving time STi and departure time Edi(the time format is hh:mm, 0<=hh<24, 0<=mm<60), Given that the arriving time must be earlier than the departure time.
Pay attention that when a group of people arrive at the restaurant as soon as a group of people leaves from the restaurant, then the arriving group can be arranged to take their seats if the seats are enough.
2 2 6 08:00 09:00 5 08:59 09:59 2 6 08:00 09:00 5 09:00 10:00
11 6
//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
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 409 Accepted Submission(s): 101
For each test case, the first line contains two integers n(0<=n<=1000), L(1<=L<=100000), which mean n gas stations in all and the distance TIANKENG can drive with the full tank respectively. The second line contains two integers sx, sy, which refers to the coordinate of the school.
The third line contains two integers ex, ey, which refers to the coordinate of TIANKENG’s home.
Then following n lines, each line contains two integer xi, yi, which refers to the coordinate of gas station.
Assuming that all the coordinates are different! All the coordinates satisfied [0,1000000].
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;
}
