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;
}