ARC 076 E - Connected?

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题意：

给出一个Ｒ×Ｃ的木板，上面有Ｎ对数字，相同数字之间连线（可以是弯的），判定是否可以不相交。

算法：

考虑对答案有影响的点对，是那些两个点都在边界上的点对，然后将矩形边展开成一条线段，在线段上判定点对投影的线段之间是否可以不相交。

代码：

#include <bits/stdc++.h>

/*
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <queue>
#include <vector>
#include <set>
*/

using namespace std;

typedef long long LL;
typedef double DB;
typedef unsigned int UI;
typedef pair<int, int> PII;

const int inf = 0x7f7f7f7f;

#define rdi() read<int>()
#define rdl() read<LL>()
#define rds(a) scanf("%s", a)
#define mk(i, j) make_pair(i, j)
#define pb push_back
#define fi first
#define se second
#define For(i, j, k) for (int i = j; i <= k; i ++)
#define Rep(i, j, k) for (int i = j; i >= k; i --)
#define Edge(i, u) for (int i = head[u]; i; i = e[i].nxt)

template<typename t> t read() {
    t x = 0; int f = 1; char c = getchar();
    while (c > '9' || c < '0') f = c == '-' ? -1 : 1 , c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + c - 48 , c = getchar();
    return x * f;
}

template<typename t> void write(t x) {
    if (x < 0){
        putchar('-'), write(-x);
        return;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + 48);
}

const int N = 2e5 + 10;
int r, c, n;
int x[N][2], y[N][2];
bool exist[N];
stack <int> stk;

int getpos(int x, int y) {
    if (!y) return x;
    if (x == r) return r + y;
    if (y == c) return r + c + r - x;
    return r + c + r + (c - y); 
}

int main() {
//  freopen(".in", "r", stdin);
//  freopen(".out", "w", stdout);
    r = rdi(), c = rdi(), n = rdi();
    for (int i = 0; i < n; i ++) x[i][0] = rdi(), y[i][0] = rdi(), x[i][1] = rdi(), y[i][1] = rdi();
    vector <PII> nums;
    for (int j, i = 0; i < n; i ++) {
        for (j = 0; j < 2; j ++) {
            if (!(x[i][j] == 0 || x[i][j] == r || y[i][j] == 0 || y[i][j] == c)) break;
        }
        if (j == 2) {
            int p = getpos(x[i][0], y[i][0]);
            int q = getpos(x[i][1], y[i][1]);
            nums.push_back(mk(p, i));
            nums.push_back(mk(q, i));
        }
    }
    sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size(); i ++) {
        int v = nums[i].second;
        if (!exist[v]) {
            stk.push(v);
            exist[v] = true;
        }
        else {
            if (stk.top() != v) return puts("NO"), 0;
            stk.pop();
            exist[v] = false;
        }           
    }
    return puts("YES"), 0;
}