Piotr‘s Ants

Problem Description

"One thing is for certain: there is no stopping them;the ants will soon be here. And I, for one, welcome our new insect overlords."Kent Brockman

Piotr likes playing with ants. He has n of them on a horizontal pole L cm long. Each ant is facing either left or right and walks at a constant speed of 1 cm/s. When two ants bump into each other, they both turn around (instantaneously) and start walking in opposite directions. Piotr knows where each of the ants starts and which direction it is facing and wants to calculate where the ants will end up T seconds from now.

Input

The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing 3 integers: L , T and n (0 ≤ n ≤ 10000). The next n lines give the locations of the n ants (measured in cm from the left end of the pole) and the direction they are facing (L or R).

Output

For each test case, output one line containing ‘Case #x:’ followed by n lines describing the locations and directions of the n ants in the same format and order as in the input. If two or more ants are at the same location, print ‘Turning’ instead of ‘L’ or ‘R’ for their direction. If an ant falls off the pole before T seconds, print ‘Fell off’ for that ant. Print an empty line after each test case.

Sample Input

2

10 1 4

1 R

5 R

3 L

10 R

10 2 3

4 R

5 L

8 R

Sample Output

Case #1:

2 Turning

6 R

2 Turning

Fell off

Case #2:

3 L

6 R

10 R

---

Solution

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int maxn=10000+5;
string name[3]={"L","Turning","R"};
//理解本题目的关键在于蚂蚁无论怎么运动，Ts后其相对位置不会改变 
struct ant
{
	int id;//输入顺序
	int p;//位置
	int d;//朝向：-1：左  0：Turning 1：右
	bool operator < (const ant& a) const
	{
		return p<a.p;
	 } 
};
ant before[maxn],after[maxn];
int order[maxn];
int main()
{
	int T;
	cin>>T;
	for(int t=0;t<T;t++)
	{
		cout<<"Case #"<<t+1<<":"<<endl;
		int L,T,n;
		cin>>L>>T>>n;
		for(int i=0;i<n;i++)
		{
			int p,d;
			char c;
			cin>>p>>c;
			d=(c=='L'?-1:1);
			before[i]=(ant){i,p,d};
			after[i]=(ant){0,p+T*d,d};//after的输入id无关紧要 
		}
		sort(before,before+n);
		for(int i=0;i<n;i++)
		order[before[i].id]=i;//order数组建立实际输入顺序->在木棍上移动实际位置的映射
		sort(after,after+n);
		for(int i=0;i<n-1;i++)
		{
			if(after[i].p==after[i+1].p)
			after[i].d=after[i+1].d=0;//修改正在碰撞蚂蚁的方向 
	    } 
	    for(int i=0;i<n;i++)
	    {
	    	int a=order[i];
			//order数组第i个对应数值（i代表输入顺序） 
			//其实就是该只 蚂蚁实际位置（体现在after数组的位置 
			if(after[a].p<0||after[a].p>L)
			cout<<"Fell off"<<endl;
			else
			cout<<after[a].p<<" "<<name[after[a].d+1]<<endl;
		}
		cout<<endl;
	}
	
	return 0;
}

Python版本　

def sort_dict(d):
    for j in range(len(d)):
        mindex=j;
        for i in range(j+1,len(d)):
            if(d[i]["pos"]<d[mindex]["pos"]):
                mindex=i
        temp=d[mindex]
        d[mindex]=d[j]
        d[j]=temp               
    
name=["L","Turning","R"]
T=int(input());
for i in range(T):
    print("Case #"+str(i+1)+":")
    s=input()
    w=s.split()
    length=int(w[0])
    time=int(w[1])
    num=int(w[2])
    before=[]
    after=[]
    for j in range(num):
        temp=dict()
        temp1=dict()
        temp1["index"]=temp["index"]=j
        s1=input()
        w1=s1.split()
        temp["pos"]=int(w1[0])
        d=w1[1]
        if(d=="L"):
            temp1["dire"]=temp["dire"]=-1
        else:
            temp1["dire"]=temp["dire"]=1
        before.append(temp)
        temp1["pos"]=temp["pos"]+time*temp["dire"]
        after.append(temp1)
    sort_dict(before)
    order=list(range(num))
    for i in range(num):
        order[before[i]["index"]]=i
    sort_dict(after)
    for i in range(num-1):
        if(after[i]["pos"]==after[i+1]["pos"]):
            after[i]["dire"]=after[i+1]["dire"]=0
    for i in range(num):
        start=order[i]
        if(after[start]["pos"]<0 or after[start]["pos"]>length):
            print("Fell off
")
        else:
            print(str(after[start]["pos"])+" "+str(name[after[start]["dire"]+1])+"
")