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

题目大意：给你一个n*m的矩阵，再给你一个小球，从(0,0)以sqrt(2)/s的速度向右上角出发，遇到边框会反弹，遇到角落就直接停止，给你一些点，问小球第一次经过这些点所需要的时间。

思路：模拟一下即可。。。注意爆int

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")
const int maxn = 5e5 + 5;
map<pair<LL, LL>, int> line;
map<pair<LL, LL>, int> ID;
vector<LL> v[maxn];
bool vis[maxn];
LL X[maxn], Y[maxn];
LL ans[maxn];
int n, m, k;
int cnt;

int get_id(LL k, LL b){
    if (ID.count(mk(k, b))) return ID[mk(k, b)];
    ID[mk(k, b)] = ++cnt;
    return cnt;
}

bool check(LL k, LL b){
    if (line.count(mk(k, b))) return true;
    line[mk(k, b)] = 1; return false;
}

void cal_time(int k, int b, LL x, LL y, LL colck){
    if (ID.count(mk(k, b)) == 0) return ;
    for (int i = 0; i < v[ID[mk(k, b)]].size(); i++){
        int pos = v[ID[mk(k, b)]][i];
        if (vis[pos]) continue;
        vis[pos] = true;
        LL tx = abs(X[pos] - x), ty = abs(Y[pos] - y);
        ans[pos] = colck + min(tx, ty);
    }
}
void solve(){
    memset(ans, -1, sizeof(ans));
    LL colck = 0;
    int ty = 1;
    int x = 0, y = 0;
    while(true){
        int nx, ny;
        if (ty == 1){
            if(check(1, y - x)) break;
            cal_time(1, y - x, x, y, colck);
            nx = n - x, ny = m - y;
            if (nx < ny) x = n, y = y + nx, ty = 2;
            if (nx > ny) x = x + ny, y = m, ty = 4;
        }
        else if (ty == 2){
            if(check(-1, y + x)) break;
            cal_time(-1, y + x, x, y, colck);
            nx = x, ny = m - y;
            if (nx < ny) x = 0, y = y + nx, ty = 1;
            if (nx > ny) x = x - ny, y = m, ty = 3;
        }
        else if (ty == 3){
            if(check(1, y - x)) break;
            cal_time(1, y - x, x, y, colck);
            nx = x, ny = y;
            if (nx < ny) x = 0, y = y - nx, ty = 4;
            if (nx > ny) x = x - ny, y = 0, ty = 2;
        }
        else if (ty == 4){
            if(check(-1, y + x)) break;
            cal_time(-1, y + x, x, y, colck);
            nx = n - x, ny = y;
            if (nx < ny) x = n, y = y - nx, ty = 3;
            if (nx > ny) x = x + ny, y = 0, ty = 1;
        }
        colck += min(nx, ny);
        if(nx == ny) break;
    }
}

int main(){
    cin >> n >> m >> k;
    for (int i = 1; i <= k; i++){
        scanf("%lld%lld", X + i, Y + i);
        int t1 = get_id(-1, X[i] + Y[i]);
        int t2 = get_id(1, Y[i] - X[i]);
        v[t1].push_back(i);
        v[t2].push_back(i);
    }
    solve();
    for (int i = 1; i <= k; i++){
        printf("%lld
", ans[i]);
    }
    return 0;
}