HDU 6136 Death Podracing (2017 Multi-Univ Training Contest 8)
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Problem
Problem Link
Idea
解题的必要前提:将剩余的所有人按照初始位置从小到大排序,重新依次编号 1 ~ n 。则显然,最早的两人相遇一定发生在某两个编号相邻的人之间(1 与 n 也视作相邻)。
故用 set 维护剩余的人的编号,当某个编号为 i 的人被淘汰后,在 set 中删除编号 i ,同时查找与 i 左相邻及右相邻的两个编号,显然此时这两个编号也又可能是在淘汰 i 之后,最快相遇的一对。
对于任意一对,结构体记录 两人的编号 x, y 与两人相遇所需时间
,压入优先队列(相遇所需时间最短的先出队列)。逐次出队,处理两人相遇,淘汰 power 较小的,同时维护 set 。当 set 中仅有两个编号时,这两人即为最晚相遇的一对。直接处理两人路程差及速度差, GCD 处理后输出。
Code
#include<bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int T, n, L;struct Point { int d, v, power;} pr[N];bool cmp(Point a, Point b) { return a.d < b.d;}struct P { int x, y; double times;} p, q;bool operator<(P a, P b) { return a.times > b.times;}int sgn(int x) { if(x == 0) return 0; return x > 0 ? 1 : -1;}double calTime(int x, int y) { int dist = pr[y].d - pr[x].d; int dv = pr[x].v - pr[y].v; if(dv < 0) { dist = L - dist; dv *= -1; } return dist * 1.0 / dv;}set<int> st;set<int>::iterator it;void solve() { it = st.begin(); int x = *it; int y = *(++it); int dist = pr[y].d - pr[x].d; int dv = pr[x].v - pr[y].v; if(dv < 0) { dist = L - dist; dv *= -1; } int GCD = __gcd(dist, dv); printf("%d/%d\n", dist/GCD, dv/GCD); }int main(){ freopen("in", "r", stdin); scanf("%d", &T); while(T-- && scanf("%d %d", &n, &L)!=EOF) { st.clear(); for(int i=0;i<n;i++) scanf("%d", &pr[i].d); for(int i=0;i<n;i++) scanf("%d", &pr[i].v); for(int i=0;i<n;i++) { pr[i].power = i; st.insert(i); } if(n == 2) { solve(); continue; } sort(pr, pr+n, cmp); priority_queue<P> que; while(!que.empty()) que.pop(); for(int i=0;i<n;i++) { p.x = i, p.y = (i+1)%n, p.times = calTime(p.x, p.y); que.push(p); } while(!que.empty()) { p = que.top(); que.pop(); if(st.find(p.x) == st.end() || st.find(p.y) == st.end()) continue; if(pr[ p.x ].power > pr[ p.y ].power) { swap(p.x, p.y); } st.erase(p.x); if(st.size() == 2) { solve(); break; } it = st.lower_bound(p.x); if(it == st.begin()) it = st.end(); it--; q.x = *it; it = st.lower_bound(p.x); if(it == st.end()) it = st.begin(); q.y = *it; if(q.x > q.y) swap(q.x, q.y); q.times = calTime(q.x, q.y); que.push(q); } }}
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