Piotr's Ants,UVa 10881

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【原题】
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】
长度为L的木棍上 有n只蚂蚁，每只蚂蚁有一个初始运动方向，两只蚂蚁相遇后立刻掉头。每秒走一个单位 ，求T秒后每只蚂蚁的顺序 ，要求按照输入时的顺序输出 。
【分析】
1.“掉头”等价于“对穿而过”，但我们要弄清楚具体的位置是哪只蚂蚁。
2. 所有蚂蚁的相对顺序是保持不变的，因此把所有目标位置从大到小排序，对应于初始状态时从左到右的每只蚂蚁。
3. 题目要求按照输入顺序输出每只蚂蚁的位置和朝向，所以我们需要一个数组order[i]来存储蚂蚁的输入序号。
4. 输出的顺序问题 就靠order数组 来解决 ，order[i]记录的是原来id是 i的现在所在顺序序列的第几位

#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("Felll off\n");else printf("%d %s\n", after[a].p, dirName[after[a].d + 1]);}printf("\n");}return 0;
}

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