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

分析：

把所有的点按照他们所在地斜线分类，斜线形如y=x+b,y=−x+b。然后模拟整个运动过程，每次都是一个斜线到另一个斜线的变换，最多有O(max(n,m))的复杂度。
比赛的时候快速想出了整个思路，并且紫色渡劫成功。欢迎读者关注Codeforces id：JIBANCANYANG。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>using namespace std;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;const int bias = int(1e5) + 13, maxn = int(2e5 + 30);struct node {    int x, y, d;};struct jibancanyang{    vector<node> A[maxn], B[maxn];    int n, m, k;    long long ans[bias];    void fun() {        scanf("%d%d%d", &n, &m, &k);        memset(ans, -1, sizeof(ans));        for (int i = 0; i < k; ++i) {            node a;            scanf("%d%d", &a.x, &a.y);            a.d = i;            A[a.y - a.x + bias].push_back(a);            B[a.x + a.y].push_back(a);        }        int s = 0, t = 0, d = 1;        long long times = 0;        while (true) {            int tt = t, ss = s;            if (d == 1) {                int y1 = t + (n - s);                if (y1 <= m) {                    s = n, t = y1;                     d = 4;                } else {                    s = s + (m - t), t = m;                    d = 2;                }                for (auto c : A[tt - ss + bias]) {                    if (ans[c.d] == -1) {                        ans[c.d] = times + abs(c.y - tt);                     }                }            }            else if (d == 2) {                int y1 = t - (n - s);                if (y1 >= 0) {                    s = n, t = y1;                     d = 3;                } else {                    s = s + t, t = 0;                    d = 1;                }                for (auto &c : B[tt + ss]) {                    if (ans[c.d] == -1) {                        ans[c.d] = times + abs(c.y - tt);                     }                }            }            else if (d == 3) {                int y1 = t - (s);                if (y1 >= 0) {                    s = 0, t = y1;                     d = 2;                } else {                    s = s - t, t = 0;                    d = 4;                }                for (auto &c : A[tt - ss + bias]) {                    if (ans[c.d] == -1) {                        ans[c.d] = times + abs(c.y - tt);                     }                }            }            else  {                int y1 = t + (s);                if (y1 <= m) {                    s = 0, t = y1;                     d = 1;                } else {                    s = s - (m - t), t = m;                    d = 3;                }                for (auto &c : B[tt + ss]) {                    if (ans[c.d] == -1) {                        ans[c.d] = times + abs(c.y - tt);                     }                }            }            times += abs(s - ss);            if (s == 0 && (t == 0 || t == m)) break;            if (s == n && (t == 0 || t == m)) break;        }        for (int i = 0; i < k; ++i) {            printf("%lld\n", ans[i]);        }    }}ac;int main(){#ifdef LOCAL    freopen("inn.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    ac.fun();    return 0;}