Covered Points Count（思维题）

C. Covered Points Count

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{nn segments\; on\; a\; coordinate\; line;\; each\; endpoint\; of\; every\; segment\; has\; integer\; coordinates.\; Some\; segments\; can\; degenerate\; to\; points.\; Segments\; can\; intersect\; with\; each\; other,\; be\; nested\; in\; each\; other\; or\; even\; coincide.}$

Your task is the following: for every $\mathrm{k\in [1..n]k\in [1..n],\; calculate\; the\; number\; of\; points\; with\; integer\; coordinates\; such\; that\; the\; number\; of\; segments\; that\; cover\; these\; points\; equals kk.\; A\; segment\; with\; endpoints lili and riri covers\; point xx if\; and\; only\; if li\le x\le rili\le x\le ri.}$

Input

The first line of the input contains one integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; number\; of\; segments.}$

The next $\mathrm{nn lines\; contain\; segments.\; The ii-th\; line\; contains\; a\; pair\; of\; integers li,rili,ri (0\le li\le ri\le 10180\le li\le ri\le 1018)\; —\; the\; endpoints\; of\; the ii-th\; segment.}$

Output

Print $\mathrm{nn space\; separated\; integers cnt1,cnt2,\dots ,cntncnt1,cnt2,\dots ,cntn,\; where cnticnti is\; equal\; to\; the\; number\; of\; points\; such\; that\; the\; number\; of\; segments\; that\; cover\; these\; points\; equals\; to ii.}$

Examples

input

Copy

3
0 3
1 3
3 8

output

Copy

6 2 1 

input

Copy

3
1 3
2 4
5 7

output

Copy

5 2 0 

Note

The picture describing the first example:

Points with coordinates $[0,4,5,6,7,8][0,4,5,6,7,8] are\; covered\; by\; one\; segment,\; points [1,2][1,2] are\; covered\; by\; two\; segments\; and\; point [3][3] is\; covered\; by\; three\; segments.$

The picture describing the second example:

Points $[1,4,5,6,7][1,4,5,6,7] are\; covered\; by\; one\; segment,\; points [2,3][2,3] are\; covered\; by\; two\; segments\; and\; there\; are\; no\; points\; covered\; by\; three\; segments.$

 

 这题坑点比较多，没有开LLwa一发 ，数组开小了又wa一发 难受 

 细节方面很多都没有注意到

　

　这题是纯思维题 把每一个区间分为两个点一个左端点一个右端点 

  进行排序一下，就出来了 

  从左往右扫一遍经历一次左端点就加一，经历一次又端点就减一 

  这个规律看图一下就出来了

  

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 4e5 + 10;
 4 typedef long long LL;
 5 struct node {
 6     LL x, y;
 7     node (LL x, LL y) : x(x), y(y) {}
 8     node () {}
 9 } qu[maxn];
10 int cmp(node a, node b) {
11     if (a.x == b.x) return a.y > b.y;
12     return a.x < b.x;
13 }
14 LL ans[maxn];
15 int main() {
16     LL n,k=0,a,b;
17     scanf("%lld", &n);
18     for (int i = 0 ; i < n ; i++) {
19         scanf("%lld%lld",&a,&b);
20         qu[k++] = node(a, 1);
21         qu[k++] = node(b + 1, -1);
22     }
23     sort(qu, qu + k, cmp);
24     LL temp = 0;
25     for(int i = 0 ; i < k-1  ; i++ ) {
26         temp += qu[i].y;
27         if (qu[i].x != qu[i + 1].x) ans[temp]+= qu[i + 1].x - qu[i].x;
28     }
29     for (int i = 1 ; i <=n ; i++)
30         printf("%lld ", ans[i]);
31     printf("\n");
32     return 0;
33 }

View Code

 

转载于:https://www.cnblogs.com/qldabiaoge/p/9267936.html