Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined)

Posted 2020-08-14

C 模拟

题意：给的是一个矩形，然后√2 的速度走，如果走到边上就正常反射，走到角上，暂停反射，我们知道要不循环要不暂停，记录走到的点最短时间

/*************************************************************************
    > File Name: c.cpp
    > Author: opas_chenxin
    > Mail: [e-mail] 
    > Created Time: 2016年10月08日 星期六 22时01分54秒
 ************************************************************************/
#include<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
struct point{
    ll x,y,d;
    point(ll x1 = 0, ll y1 = 0, ll d1 = 0) {
        x = x1; y = y1; d = d1;
    }
     bool operator < (point const&A) const{
         if(x != A.x) return x < A.x;
         if(y != A.y) return y < A.y;
         return d < A.d;
    }
    //bool operator == (point &A){
    //    return x == A.x && y == A.y && d == A.d;
    //}
};
map<point,ll> G;
ll n,m,k;
ll edg_w[4]={0,0,0,0};
bool GetNextPoint(int loc, point A, ll &t, point &B, int fu) {
        ll b = A.y - A.d * A.x;
        if(loc&1) {
            B.y = edg_w[loc];
            B.x = (B.y - b) * A.d;
            t = abs(A.y - edg_w[loc]);
        }    else {
            B.x = edg_w[loc];
            B.y = B.x * A.d + b;
            t = abs(A.x - edg_w[loc]); 
        }
        B.d = A.d * fu; 
        if(B.x < 0 || B.x > n || B.y < 0 || B.y > m || (B.x == A.x && B.y == A.y) )
            return false;
        return true;
}
point GetNextLoc(point &A, ll &t ) {
    point next_loc;
    ll tt;
    for(int i = 0; i < 4; ++ i) {
        if(GetNextPoint(i, A, tt, next_loc, -1)) {
            t += tt;
            return next_loc;
        }
    }
    return next_loc;
}
bool JudEnd(point &A){
    if(A.x == 0 && A.y == 0) return true;
    if(A.x == 0 && A.y == m) return true;
    if(A.x == n && A.y == 0) return true;
    if(A.x == n && A.y == m) return true;
    return false;
}
ll GetAns(point &A){
    ll t=0;
    ll ans = -1;
    for(int i = 0; i < 2; ++ i) {
        A.d = -A.d;
        for(int j = 0; j < 4; ++ j) {
            point B;
            if(GetNextPoint(j, A, t, B, 1)) {
                if(G.count(B) > 0) {
                    if(ans == -1) 
                        ans = G[B]+t;
                    else
                        ans = min(G[B]+t, ans); 
                }
            }
        }
    }
    if(ans != -1) ans = ans -1;
    return ans;
}
int main() {
    freopen("in", "r", stdin);
    while( cin>>n>>m>>k ) {
        edg_w[2] = n;
        edg_w[3] = m;
        G.clear();
        point now = point(0,0,1);
        ll t = 1;
        G.insert(pair<point,ll>(now,t));
        while(true) {
            point next_loc = GetNextLoc(now, t);
            if(JudEnd(next_loc)) break;
            if(G.count(next_loc) > 0) break;
            else G.insert(pair<point,ll>(next_loc,t));
            now = next_loc;
            G.insert(pair<point,ll>(now,t));
        }
        for(int i = 0; i < k; ++ i) {
            ll x, y;
            cin>>x>>y;
            point A = point(x, y, 1);
            cout<<GetAns(A)<<endl;
        }
    }
    return 0;
}

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