D. Rooter's Song
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
- Vertical: stands at (xi, 0), moves in positive y direction (upwards);
- Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 ≤ n ≤ 100 000, 2 ≤ w, h ≤ 100 000) — the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 ≤ gi ≤ 2, 1 ≤ pi ≤ 99 999, 0 ≤ ti ≤ 100 000), describing a dancer's group gi (gi = 1 — vertical, gi = 2 — horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 ≤ xi ≤ w - 1 and 1 ≤ yi ≤ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) — the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1 Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6 Input
3 2 3
1 1 2
2 1 1
1 1 5 Output
1 3
2 1
1 3 Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
In the second example, no dancers collide.
【題解】
不（題）難（解）發（上）現（說）兩個點能夠相碰當且僅當p - t相等
于是我們可以按照p - t分組，發現兩個點相碰可以看做穿過去。那麼
最終位置與原始位置有什麼對應關系呢？
我們把每個點的路徑化成直線，交點意味着碰撞。我們發現一個點的行走軌迹是
階梯狀的，手畫一下不難發現：左上->左下->右下方向的點對應左上->右上->
右下的點
于是兩次間接排序即可
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <cstring>
5 #include <algorithm>
6 #define max(a, b) ((a) > (b) ? (a) : (b))
7 #define min(a, b) ((a) < (b) ? (a) : (b))
8
9 const int INF = 0x3f3f3f3f;
10 const int MAXN = 1000000 + 10;
11
12 inline void read(int &x)
13 {
14 x = 0;char ch = getchar(), c = ch;
15 while(ch < '0' || ch > '9')c = ch, ch = getchar();
16 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
17 if(c == '-')x = -x;
18 }
19
20 inline int abs(int a)
21 {
22 return a < 0 ? -1*a : a;
23 }
24
25 struct Node
26 {
27 int g, p, t, x, y;
28 }node[MAXN],ans[MAXN];
29
30 int n,w,h,cnt[MAXN],cnt2[MAXN];
31
32 int cmp(int c, int d)
33 {
34 Node a = node[c], b = node[d];
35 return (a.p - a.t == b.p - b.t) ? ( (a.y == b.y) ? (a.x > b.x) : (a.y < b.y)) : (a.p - a.t < b.p - b.t);
36 }
37
38 int cmp2(int c, int d)
39 {
40 Node a = node[c], b = node[d];
41 return (a.p - a.t == b.p - b.t) ? ((a.x == b.x) ? (a.y < b.y) : (a.x > b.x)) : (a.p - a.t < b.p - b.t);
42 }
43
44 int main()
45 {
46 read(n), read(w), read(h);
47 for(register int i = 1;i <= n;++ i)
48 {
49 read(node[i].g), read(node[i].p), read(node[i].t);
50 if(node[i].g == 1) node[i].y = -1, node[i].x = node[i].p;
51 else node[i].y = node[i].p, node[i].x = -1;
52 cnt[i] = cnt2[i] = i;
53 }
54 std::sort(cnt + 1, cnt + 1 + n, cmp);
55 for(register int i = 1;i <= n;++ i)
56 if(node[i].x != -1)node[i].y = h;
57 else node[i].x = w;
58 std::sort(cnt2 + 1, cnt2 + 1 + n, cmp2);
59 for(register int i = 1;i <= n;++ i)
60 ans[cnt[i]] = node[cnt2[i]];
61 for(register int i = 1;i <= n;++ i)
62 printf("%d %d
", ans[i].x, ans[i].y);
63 return 0;
64 } 849D