Description

3
3 2 4
4 3 4
3 6 4

2 2 2

一個人在t[i]之前不存在, 在走到f[i]之後消失.

Solution

code

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<queue>
 7 #include<vector>
 8 #include<stack>
 9
 
 #include<set>
10 #include<deque>
11 #include<map>
12 using namespace std;
13 
14 template <typename T> void read(T &x) {
15     x = 0; int f = 1; char c;
16     for (c = getchar(); c < '0' || c > '9'; c = getchar()) if (c == '-') f = -f;
17     for (; c >= '0' && c <= '
 
9'; c = getchar()) x = 10 * x + c - '0' ;
18     x *= f;
19 }
20 template <typename T> void write(T x){
21     if (x < 0) putchar('-'), x = -x;
22     if (x > 9) write(x / 10);
23     putchar(x % 10 + '0');
24 }
25 template <typename T> void writeln(T x) { write(x); putchar('\n'); }
26 template <typename T> void
 
 writesn(T x) { write(x); putchar(' '); }
27 
28 #define int long long
29 #define inf 100000
30 #define next net
31 #define P 9999991
32 #define N 1010
33 #define mid ((l+r)>>1)
34 #define lson (o<<1)
35 #define rson (o<<1|1)
36 #define R register
37 #define debug puts("zxt")
38 
39 int n, ans[N ];
40 struct node{
41     double k, b;
42     int st, ed, num ;
43 }e[N ];
44 inline bool cmp(node a, node b)
45 {
46     return a.st < b.st;
47 }
48 inline bool check(int x, int y)
49 {
50     if(e[x].ed < e[y].st) return false;
51     double t1 = (e[x].k * 1.0 * e[y].st + e[x].b - e[y].k * 1.0 * e[y].st - e[y].b);
52     double t2 = (e[x].k * 1.0 * min(e[x].ed, e[y].ed) + e[x].b - e[y].k * 1.0 * min(e[x].ed, e[y].ed) - e[y].b);
53     return t1 * t2 <= 0;
54 }
55 signed main()
56 {
57     //freopen("walk.in","r",stdin);
58     //freopen("walk.out","w",stdout);
59     read(n);
60     for(R int i = 1, t, s, f; i <= n; i++)
61     {
62         read(t); read(s); read(f);
63         e[i].num = i;
64         e[i].st = t;
65         e[i].ed = t + abs(f - s);
66         if(f - s < 0) e[i].k = -1.0;
67         else e[i].k = 1.0;
68         e[i].b = 1.0 * s - 1.0 * t * e[i].k;
69     }
70     sort(e + 1, e + n + 1, cmp);
71     for(R int i = 1; i <= n; i++)
72         for(R int j = i + 1; j <= n; j++)
73         {
74             if(check(i, j)) ans[e[i].num ]++, ans[e[j].num ]++;
75         }
76     for(R int i = 1; i <= n; i++) writesn(ans[i]);
77     return 0;
78 }