UVa10881

题目名称：Piotr's Ants

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=20&page=show_problem&problem=1822

Problem D
Piotr's Ants
Time Limit: 2 seconds

"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 nlines 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 InputSample Output

210 1 41 R5 R3 L10 R10 2 34 R5 L8 R

Case #1:2 Turning6 R2 TurningFell offCase #2:3 L6 R10 R

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题意：有n只蚂蚁在L cm的木棍上爬行，1 cm/s，每只蚂蚁给出初设位置和爬的方向（left or right），问T时间后每只蚂蚁的位置及朝向(R ,L ,如果碰头的就Turning,掉下去的就Fell off)

思路:从远处看可以把蚂蚁看成小黑点，那样碰头就可以看成直接穿过去，这样就只要标记哪只蚂蚁在哪里就行了

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;char z[][10]={"L","Turning","R"};struct Ant{    int id,p;    int turn;    bool operator < (const Ant j) const    {        return p<j.p;    }};int main(){    int t,n,N,L,a[10005],d[10005],f=1;    Ant x[10005],y[10005];    while(~scanf("%d",&N))    {        for(int i=0;i<N;i++)        {            scanf("%d%d%d",&L,&t,&n);            for(int j=0;j<n;j++)            {                int p,d;                char c;                scanf("%d %c",&p,&c);                d=(c=='L'?-1:1);                x[j]=(Ant){j,p,d};                y[j]=(Ant){0,p+t*d,d};            }            sort(x,x+n);            for(int j=0;j<n;j++)                d[x[j].id]=j; //记录原来第几个输入的排序后在哪            sort(y,y+n);            for(int j=0;j<n;j++)            {                if(y[j].p==y[j+1].p)                    y[j].turn=y[j+1].turn=0;            }            printf("Case #%d:\n",f++);            for(int j=0;j<n;j++)            {                int a=d[j];                if(y[a].p<0||y[a].p>L)                    printf("Fell off\n");                else                {                    printf("%d %s\n",y[a].p,z[y[a].turn+1]);                }            }            printf("\n");        }    }    return 0;}