UVa 10881Piotr's Ants【思维】

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题目链接

对于每个蚂蚁来说，他相对于其他蚂蚁的位置是固定的，赶超前面或者被后面的赶超都是要回头的，所以，假设木棒无限长，位于第一位的永远在第一位，第二位的永远在第二位。同时每两个碰头的蚂蚁其实就是类似于交换了身份继续前进。所以，先不管他们的碰头，就让每个沿着路线继续走下去。由前面的可以知道，每个蚂蚁的最终状态的位置排列的顺序就是一开始蚂蚁的编号。

#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef unsigned long long UL;//typedef __int64 Int;typedef pair<int, int> PAI;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const int MOD = 1e9 + 7;const int MAXN = 10000 + 10;struct ant {    int dic;    int s, e;    int id;} data[MAXN];int f[MAXN];char str[][10] = {"L", "Turning", "R", "Fell off"};bool cmp(ant x, ant y) {return x.e < y.e;}bool cmp1(ant x, ant y) {return x.s < y.s;}int main() {    int T;    char s[10];    scanf("%d", &T);    int Kcase = 0;    while (T--) {        int L, t, N;        scanf("%d%d%d", &L, &t, &N);        for (int i = 0; i < N; i++) {            scanf("%d%s", &data[i].s, s);            data[i].dic = (s[0] == 'R'? 1: -1);            data[i].id = i;            data[i].e = data[i].s + data[i].dic*t;        }        sort(data, data + N, cmp1);        for (int i = 0; i < N; i++) f[data[i].id] = i;        sort(data, data + N, cmp);        for (int i = 0; i < N; i++) {            if (data[i].e < 0 || data[i].e > L) data[i].dic = 2;            else if (i != 0 && data[i].e == data[i - 1].e) data[i].dic = data[i - 1].dic = 0;        }        printf("Case #%d:\n", ++Kcase);        for (int i = 0; i < N; i++) {            int u = f[i];            int v = data[u].dic;            if (v == 2) printf("Fell off\n");            else printf("%d %s\n", data[u].e, str[v+1]);        }        printf("\n");    }    return 0;}

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