[ABC273D] LRUD Instructions 题解

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[ABC273D] LRUD Instructions 题解

很好的一道大模拟，使我爆 $0$。

思路解析

大模拟，我们只需要用一个 $x,y$ 表示我们当前的位置，而对于每一个移动，我们就判断在当前移动方向上离当前点最近的点，若该点在当前行进路线上，则把当前位置设为该点前面的一个。

其中判断在当前移动方向上离当前点最近的点可以使用离散化，然后使用二分法查找第一个大于当前点的障碍物，而若是朝负方向移动，则可以修改 lower_bound 里的迭代器起终点和 cmp 函数，这样就可以用 lower_bound 实现反向二分。

PS：考试时能用 vector 就千万不要用 set！！！常数奇大无比！！ 挂分记录

code

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fir first
#define sec second
int h, w, n;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int sx, sy;
map<int, vector<int> > mph, mpl;
map<char, int> mpc;
int q;
void init() {
	mpc['U'] = 0;
	mpc['D'] = 1;
	mpc['L'] = 2;
	mpc['R'] = 3;
}
signed main() {
	init();
	cin >> h >> w >> sx >> sy;
	cin >> n;
	for(int i = 1, r, c; i <= n; i++) {
		cin >> r >> c;
		mph[r].push_back(c);
		mpl[c].push_back(r);
	}
	for(auto &it : mph) {
		it.sec.push_back(0);
		it.sec.push_back(2e9);
		sort(it.sec.begin(), it.sec.end());
	}
	for(auto &it : mpl) {
		it.sec.push_back(0);
		it.sec.push_back(2e9);
		sort(it.sec.begin(), it.sec.end());
	}
	cin >> q;
	int x = sx, y = sy;
	while(q--) {
		char ch;
		int l;
		cin >> ch >> l;
		int d = mpc[ch];
		int nx = x + dx[d] * l;
		nx = max(1ll, nx);
		nx = min(nx, h);
		int ny = y + dy[d] * l;
		ny = max(1ll, ny);
		ny = min(ny, w);
		if(d == 0) {
			if(!mpl[ny].empty()) {
				int temp = *lower_bound(mpl[ny].rbegin(), mpl[ny].rend(), x, [](int xx, int yy) {	//lambda表达式
					return xx > yy;
				});
				if(temp >= nx && temp <= x) {
					x = temp + 1;
				}
				else {
					x = nx;
				}
			}
			else {
				x = nx;
			}
		}
		else if(d == 1) {
			if(!mpl[ny].empty()) {
				int temp = *lower_bound(mpl[ny].begin(), mpl[ny].end(), x);
				if(temp >= x && temp <= nx) {
					x = temp - 1;
				}
				else {
					x = nx;
				}
			}
			else {
				x = nx;
			}
		}
		else if(d == 2) {
			if(!mph[nx].empty()) {
				int temp = *lower_bound(mph[nx].rbegin(), mph[nx].rend(), y, [](int xx, int yy) {
					return xx > yy;
				});
				if(temp >= ny && temp <= y) {
					y = temp + 1;
				}
				else {
					y = ny;
				}
			}
			else {
				y = ny;
			}
		}
		else if(d == 3) {
			if(!mph[nx].empty()) {
				int temp = *lower_bound(mph[nx].begin(), mph[nx].end(), y);
				if(temp >= y && temp <= ny) {
					y = temp - 1;
				}
				else {
					y = ny;
				}
			}
			else {
				y = ny;
			}
		}
		cout << x << " " << y << "\n";
	}
	return 0;
}