Description
There is given the series of n closed intervals [ai; bi], where i=1,2,...,n. The sum of those intervals may be represented as a sum of closed pairwise non−intersecting intervals. The task is to find such representation with the minimal number of intervals. The intervals of this representation should be written in the output file in acceding order. We say that the intervals [a; b] and [c; d] are in ascending order if, and only if a <= b < c <= d.
Task
Write a program which:
reads from the std input the description of the series of intervals,
computes pairwise non−intersecting intervals satisfying the conditions given above,
writes the computed intervals in ascending order into std output
Input
In the first line of input there is one integer n, 3 <= n <= 50000. This is the number of intervals. In the (i+1)−st line, 1 <= i <= n, there is a description of the interval [ai; bi] in the form of two integers ai and bi separated by a single space, which are respectively the beginning and the end of the interval,1 <= ai <= bi <= 1000000.
Output
The output should contain descriptions of all computed pairwise non−intersecting intervals. In each line should be written a description of one interval. It should be composed of two integers, separated by a single space, the beginning and the end of the interval respectively. The intervals should be written into the output in ascending order.
Sample Input
5
5 6
1 4
10 10
6 9
8 10 Sample Output
1 4
5 10 Source
POI VIII Stage 1 Problem 2 正解：貪心+排序 解題報告： 水題。區間覆寫，排一遍序即可。
1 //It is made by jump~
2 #include <iostream>
3 #include <cstdlib>
4 #include <cstring>
5 #include <cstdio>
6 #include <cmath>
7 #include <algorithm>
8 #include <ctime>
9 #include <vector>
10 #include <queue>
11 #include <map>
12 #ifdef WIN32
13 #define OT "%I64d"
14 #else
15 #define OT "%lld"
16 #endif
17 using namespace std;
18 typedef long long LL;
19 const int MAXN = 50011;
20 int n;
21
22 struct seq{
23 int l,r;
24 }a[MAXN];
25
26 inline bool cmp(seq q,seq qq){ if(q.l==qq.l) return q.r<qq.r; return q.l<qq.l; }
27
28 inline int getint()
29 {
30 int w=0,q=0;
31 char c=getchar();
32 while((c<'0' || c>'9') && c!='-') c=getchar();
33 if (c=='-') q=1, c=getchar();
34 while (c>='0' && c<='9') w=w*10+c-'0', c=getchar();
35 return q ? -w : w;
36 }
37
38
39 inline void solve(){
40 n=getint();
41 for(int i=1;i<=n;i++) {
42 a[i].l=getint(); a[i].r=getint();
43 }
44 sort(a+1,a+n+1,cmp);
45 int nowl=a[1].l,nowr=a[1].r;
46 for(int i=2;i<=n;i++) {
47 if(a[i].l>nowr) {
48 printf("%d %d\n",nowl,nowr);
49 nowl=a[i].l; nowr=a[i].r;
50 }
51 else if(a[i].r>nowr) nowr=a[i].r;
52 }
53
54 printf("%d %d",nowl,nowr);
55 }
56
57 int main()
58 {
59 solve();
60 return 0;
61 } 轉載于:https://www.cnblogs.com/ljh2000-jump/p/5605087.html