Codeforces 848B Rooter's Song(分类+模拟)

题目链接 Rooter's Song

题意  有n个舞者站在x轴上或y轴上,每个人有不同的出发时间。x轴上的舞者垂直x轴正方向移动,y轴上的舞者垂直y轴正方向移动。

当x轴的舞者和y轴的舞者相遇时,他们会互换运动轨迹。求每个舞者的最后位置。

 

把所有会发生碰撞的舞者塞到一起,按照坐标大小升序排序。

对于某个在x轴上的舞者, 计算他右边的舞者个数,和他上面的舞者个数(即y轴会和他碰撞的所有舞者个数)

对于某个在y轴上的舞者, 计算他上面的舞者个数,和他右边的舞者个数(即x轴会和他碰撞的所有舞者个数)

然后根据这些信息计算出每个舞者的碰撞次数,再求出每个舞者最后的位置。

时间复杂度$O(Mlogn)$    $(M = max(w, h))$

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

typedef long long LL;

const int N = 4e5 + 10;
const int d = 1e5;

struct node{
	int pos, id;
	friend bool operator < (const node &a, const node &b){
		return a.pos < b.pos;
	}
};

int n, w, h, op, p, t, mx, s;
vector <node> X[N], Y[N];
int flag[N], ans[N], b[N], c[N];

int main(){

	scanf("%d%d%d", &n, &w, &h);
	mx = 0;

	rep(i, 1, n){
		scanf("%d%d%d", &op, &p, &t);
		b[i] = op, c[i] = p;
		if (op == 1) X[p - t + d].push_back({p, i});
		else Y[p - t + d].push_back({p, i});
		mx = max(mx, p - t + d);
	}

	rep(i, 0, mx){
		sort(X[i].begin(), X[i].end());
		sort(Y[i].begin(), Y[i].end());
	}

	rep(i, 0, mx){
		s = (int)X[i].size();
		t = (int)Y[i].size();

		if (s == 0 || t == 0) continue;
		
		rep(j, 0, s - 1){
			int nx  = s - j - 1, ny = t;
			int num;
			if (nx < ny) num = 2 * nx + 1;
			else num = ny * 2;
			if (num & 1) flag[X[i][j].id] = 2; else flag[X[i][j].id] = 1;
			if (flag[X[i][j].id] == 1) ans[X[i][j].id] = X[i][j + num / 2].pos;
			else ans[X[i][j].id] = Y[i][num / 2].pos;
		}

		rep(j, 0, t - 1){
			int nx  = s, ny = t - j - 1;
			int num;
			if (ny < nx) num = 2 * ny + 1;
			else num = nx * 2;
			if (num & 1) flag[Y[i][j].id] = 1; else flag[Y[i][j].id] = 2;
			if (flag[Y[i][j].id] == 2) ans[Y[i][j].id] = Y[i][j + num / 2].pos;
			else ans[Y[i][j].id] = X[i][num / 2].pos;
			
		}
	}

	
	rep(i, 1, n) if (!flag[i]){ flag[i] = b[i];  ans[i] = c[i];}
	rep(i, 1, n) if (flag[i] == 1) printf("%d %d\n", ans[i], h);
			else printf("%d %d\n", w, ans[i]);
	return 0;
}

  

posted @ 2017-09-09 18:50  cxhscst2  阅读(212)  评论(0编辑  收藏  举报