题目:

​出题人在\(x\)轴上放置了\(n\)个正在移动的炸弹,第iii个炸弹的初始位置为\(x[i]\),速度为\(v[i]\),当两颗炸弹相遇时会发生爆炸,导致这两颗炸弹消失。在经历了\(10^{100000}\)秒后,出题人想知道最后还剩下几颗炸弹,以及它们的编号。(数据保证不会有三个及以上的炸弹同时相遇)

分析:

​由于炸弹运动时间可以看做无限长,所以所有有相对运动迹象的炸弹都会爆炸,我们可以将所有相邻的炸弹的相遇时间扔进优先队列(对于炸弹的3*3种情况讨论),用set维护下标删除炸弹即可。

实现:

#include     using namespace std;#define rep(i, a, n) for(int i = a; i < n; i++)#define all(x) x.begin(), x.end()#define pb push_back#define ios ios::sync_with_stdio(false);cin.tie(0);#define debug(x)    cout << x << endl;#define SZ(x)    (int)x.size()typedef long long LL;typedef unsigned long long ULL;typedef pair PII;const double inf = 1e18;void read(int &x) {int s = 0, f = 1; char ch = getchar(); while(!isdigit(ch)) {f = (ch == '-' ? -1 : f); ch = getchar();} while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = getchar();} x = s * f;}const int N = 100005;int n;int vis[N];struct node {    int x, v, id;    bool operator < (const node& aa) const {        return x < aa.x;    }} a[N];struct Node { //优先队列维护碰撞时间最短的两个炸弹    int u, v;    double time;    bool operator  aa.time;    }};priority_queue q;set s;double calc(int idx1, int idx2) //计算相邻的炸弹的碰撞时间{    double dis = abs(a[idx1].x - a[idx2].x) * 1.0;    int v1 = a[idx1].v, v2 = a[idx2].v;    if(v1 == v2) //都为0 或者速度相同        return inf;    if(v1 == 0)    {        if(v2  0)            return dis / abs(v1);        else             return inf;    }    if(v1  0)        return inf;    if(v1 > 0 && v2 < 0)        return dis / (abs(v1) + abs(v2));        if(v1 < 0 && v2  abs(v1))            return dis / (abs(v2) - abs(v1));        else               return inf;    }    if(v1 > 0 && v2 > 0)    {        if(abs(v1) > abs(v2))            return dis / (abs(v1) - abs(v2));        else             return inf;    }    return inf;}signed main(){    cin >> n;    for(int i = 0; i > a[i].x;        a[i].id = i;     }    rep(i, 1, n + 1)        cin >> a[i].v;    sort(a + 1, a + n + 1);    for(int i = 2; i <= n; i ++)        q.push({i - 1, i, calc(i - 1, i)});    while(q.size())    {        auto tmp = q.top();        q.pop();        if(tmp.time == inf) break;        if(vis[tmp.u] || vis[tmp.v])    continue;        vis[tmp.u] = 1;        vis[tmp.v] = 1;        int l, r;        auto iter = s.find(tmp.u);        l = *(--iter);        iter ++;        s.erase(iter);        iter = s.find(tmp.v);        r = *(++iter);        iter --;        s.erase(iter);        if(l == 0 || r == n + 1)    continue;        q.push({l, r, calc(l, r)});    }    vector res;    for(int x : s)        if(x != 0 && x != n + 1)            res.pb(a[x].id);        sort(all(res));    cout << SZ(res) << '\n';    for(int x : res)        cout << x << '\n';}