2013
12-09

# Taxi Cab Scheme

Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible, there is also a need to schedule all the taxi rides which have been booked in advance. Given a list of all booked taxi rides for the next day, you want to minimise the number of cabs needed to carry out all of the rides.

For the sake of simplicity, we model a city as a rectangular grid. An address in the city is denoted by two integers: the street and avenue number. The time needed to get from the address a, b to c, d by taxi is |a – c| + |b – d| minutes. A cab may carry out a booked ride if it is its first ride of the day, or if it can get to the source address of the new ride from its latest, at least one minute before the new ride’s scheduled departure. Note that some rides may end after midnight.

On the first line of the input is a single positive integer N, telling the number of test scenarios to follow. Each scenario begins with a line containing an integer M, 0 < M < 500, being the number of booked taxi rides. The following M lines contain the rides. Each ride is described by a departure time on the format hh:mm (ranging from 00:00 to 23:59), two integers a b that are the coordinates of the source address and two integers c d that are the coordinates of the destination address. All coordinates are at least 0 and strictly smaller than 200. The booked rides in each scenario are sorted in order of increasing departure time.

For each scenario, output one line containing the minimum number of cabs required to carry out all the booked taxi rides.

2
2
08:00 10 11 9 16
08:07 9 16 10 11
2
08:00 10 11 9 16
08:06 9 16 10 11

1
2

按照题目的要求，要选择最少的taxi来完成这些任务。显然在上面这个例子中，需要安排2taxi。结合这个图，可以把题目的要求转化为找出最少的路径条数，使得这些路径覆盖途中所有的边，例如可以选择2条路径1->3->41->2就可以覆盖所有的边。也可以选择1->3->42（因为2作为初始站，不需要由1转移过来）。对于一条连续的路径vi1->vi2->…vik由于这条路径上的任务实际上是由一辆taxi来完成的，可以吧这条路径退化成两个点vi1->vik。有了这两步建图的步骤以后，问题的求解就可以变为找出顶点集的一个最小子集，使这个顶点子集覆盖所有的边（每条边都至少和一个顶点集的顶点相连）。这个问题就是图的最小点覆盖。再看这张图，还有一些性质能够让我们更好地求出最小点覆盖。这个图是一个有向无环图，没有自环，就可以拆点，把原先建的图变成一张二分图。

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"math.h"
int n,m;
char time[10];
struct node
{
int x1,x2,y1,y2;
int t,sum;
}aa[501];
int fun(int x1,int y1,int x2,int y2)
{
return abs(x1-x2)+abs(y1-y2);
}
int dfs(int k)
{
int i;
for(i=0;i<m;i++)
{
if(map[k][i]&&!v[i])
{
v[i]=1;
{
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,ans;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%s",time);
scanf("%d%d%d%d",&aa[i].x1,&aa[i].y1,&aa[i].x2,&aa[i].y2);
aa[i].t=(time[0]-'0')*10*60+(time[1]-'0')*60
+(time[3]-'0')*10+(time[4]-'0');
aa[i].sum=fun(aa[i].x1,aa[i].y1,aa[i].x2,aa[i].y2);
}
memset(map,0,sizeof(map));
for(i=0;i<m-1;i++)
{
for(j=i+1;j<m;j++)
{
if((aa[i].t+aa[i].sum+fun(aa[i].x2,aa[i].y2,aa[j].x1,aa[j].y1))<aa[j].t)
map[i][j]=1;
}
}
ans=0;
for(i=0;i<m;i++)
{
memset(v,0,sizeof(v));
ans+=dfs(i);
}
printf("%d\n",m-ans);
}
return 0;
}

1. #include <stdio.h>
int main()
{
int n,p,t[100]={1};
for(int i=1;i<100;i++)
t =i;
while(scanf("%d",&n)&&n!=0){
if(n==1)
printf("Printing order for 1 pages:nSheet 1, front: Blank, 1n");
else {
if(n%4) p=n/4+1;
else p=n/4;
int q=4*p;
printf("Printing order for %d pages:n",n);
for(int i=0;i<p;i++){
printf("Sheet %d, front: ",i+1);
if(q>n) {printf("Blank, %dn",t[2*i+1]);}
else {printf("%d, %dn",q,t[2*i+1]);}
q–;//打印表前
printf("Sheet %d, back : ",i+1);
if(q>n) {printf("%d, Blankn",t[2*i+2]);}
else {printf("%d, %dn",t[2*i+2],q);}
q–;//打印表后
}
}
}
return 0;
}

2. 很高兴你会喜欢这个网站。目前还没有一个开发团队，网站是我一个人在维护，都是用的开源系统，也没有太多需要开发的部分，主要是内容整理。非常感谢你的关注。

3. “再把所有不和该节点相邻的节点着相同的颜色”，程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的

4. 你的理解应该是：即使主持人拿走一个箱子对结果没有影响。这样想，主持人拿走的箱子只是没有影响到你初始选择的那个箱子中有奖品的概率，但是改变了其余两个箱子的概率分布。由 1/3,1/3 变成了 0, 2/3