标签：
| Merge Intervals | ||||||
|
||||||
| Description | ||||||
|
Given a collection of intervals, merge all overlapping intervals. For example, Given: [1,3],[2,6],[8,10],[15,18]; after meger: [1,6],[8,10],[15,18]. |
||||||
| Input | ||||||
|
There are multiple test cases. For each test case: Line 1: This line contains an integer n indicating the number of intervals. Line 2..n+1: Each line contains a pair of integers a and b, indicating the interval [a, b]. 1<=n<=100,000 1<=a< = b<=1,000,000,000 |
||||||
| Output | ||||||
|
Output one line, contains the intervals separated by space after the merge. Output the intervals in lexicographically smaller way. In other words, if we output [ai,bi] before [aj,bj], there must be ai < aj or ai=aj and bi <= bj. For more details, referring to the sample. |
||||||
| Sample Input | ||||||
4 1 3 2 6 8 10 15 18 5 37 65 29 43 30 42 31 87 32 86 |
||||||
| Sample Output | ||||||
[1,6] [8,10] [15,18] [29,87] |
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> using namespace std; struct node { int left; int right; } a[100001],b[100001]; bool cmp(node x,node y) { if(x.left!=y.left) { return x.left<y.left; } return x.right<y.right; } int main() { int n; while(~scanf("%d",&n)) { int i; for(i=0;i<n;i++) { scanf("%d%d",&a[i].left,&a[i].right); } sort(a,a+n,cmp); int k=0; node p=a[0]; for(i=1;i<n;i++) { if(a[i].left<=p.right&&a[i].right>p.right) { p.right=a[i].right; } if(a[i].left>p.right) { b[k++]=p; p=a[i]; } } b[k++]=p; for(i=0;i<k;i++) { printf("%c%d%c%d%c%c",‘[‘,b[i].left,‘,‘,b[i].right,‘]‘,i==k-1? ‘\n‘:‘ ‘); } } return 0; }
标签：
原文地址：http://www.cnblogs.com/beige1315402725/p/5014667.html
