uva 10881(ants)

题目：
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

题意：
一根长度为L的木棍上有n只蚂蚁，初始状态时每只蚂蚁要么朝左要么朝右爬，求T秒后每只蚂蚁的位置和状态。

ac代码：

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const char dirname[][10]={"L","Turning","R"};const int maxn=10005;struct ant{    int id;    int pos;    int dir;    bool operator < (const ant& a) const {    return pos < a.pos;  }}before[maxn],after[maxn];bool cmp(ant a,ant b){    return a.id<b.id;}int main(){    //freopen("test.txt","r",stdin);    int kase;    scanf("%d",&kase);    for(int cas=1;cas<=kase;cas++)    {        int l,t,n;        cin >> l >> t >> n;        for(int i=1;i<=n;i++)        {            char c;            int pos,d;            scanf("%d %c",&pos,&c);            d=c=='L'? -1:1;            before[i]=(ant){i,pos,d};            after[i]=(ant){0,pos+t*d,d};        }        sort(before+1,before+n+1);        sort(after+1,after+n+1);        for(int i=1;i<=n;i++)        after[i].id=before[i].id;        for(int i=1;i<n;i++)        {            if(after[i].pos==after[i+1].pos)            after[i].dir=after[i+1].dir=0;        }        sort(after+1,after+n+1,cmp);        printf("Case #%d:\n",cas);        for(int i=1;i<=n;i++)        {            if(after[i].pos<0||after[i].pos>l)            {                printf("Fell off\n");                continue;            }            printf("%d %s\n",after[i].pos,dirname[after[i].dir+1]);        }        printf("\n");    }    return 0;}

刘汝佳老师版：

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 10000 + 5;struct Ant {  int id; // 输入顺序  int p;  // 位置  int d;  // 朝向。 -1: 左; 0:转身中; 1:右  bool operator < (const Ant& a) const {    return p < a.p;  }} before[maxn], after[maxn];const char dirName[][10] = {"L", "Turning", "R"};int order[maxn]; // 输入的第i只蚂蚁是终态中的左数第order[i]只蚂蚁int main() {  int K;  scanf("%d", &K);  for(int kase = 1; kase <= K; kase++)   {    int L, T, n;    printf("Case #%d:\n", kase);    scanf("%d%d%d", &L, &T, &n);    for(int i = 0; i < n; i++) {      int p, d;      char c;      scanf("%d %c", &p, &c);      d = (c == 'L' ? -1 : 1);      before[i] = (Ant){i, p, d};      after[i] = (Ant){0, p+T*d, d}; // 这里的id是未知的    }    // 计算order数组    sort(before, before+n);    for(int i = 0; i < n; i++)      order[before[i].id] = i;    // 计算终态    sort(after, after+n);        for(int i = 0; i < n-1; i++) // 修改碰撞中的蚂蚁的方向      if(after[i].p == after[i+1].p) after[i].d = after[i+1].d = 0;    // 输出结果    for(int i = 0; i < n; i++) {      int a = order[i];       if(after[a].p < 0 || after[a].p > L) printf("Fell off\n");      else printf("%d %s\n", after[a].p, dirName[after[a].d+1]);    }    printf("\n");  }  return 0;}