POJ 1089 Intervals

 

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

55 61 410 106 98 10

Sample Output

1 45 10

思路：

将区间按照x的从小到大排序，然后顺次搜索，更新将要输出的区间的终点，直到当前区间的起点比之前要输出的区间的终点更大，将之前的区间输出。更新起点终点。

代码：

#include <stdio.h>#include <stdlib.h>#include <iostream>using namespace std;struct Interval{int beginning;int ending;};int mycompare(const void *elem1, const void* elem2){Interval *p1 = (Interval *)elem1;Interval *p2 = (Interval *)elem2;if (p1->beginning == p2->beginning)return p2->ending - p1->ending;return p1->beginning - p2->beginning;}int main(){int n;cin >> n;Interval interval[50000];for (int i = 0; i < n; i++)cin >> interval[i].beginning >> interval[i].ending;qsort(interval, n, sizeof(Interval), mycompare);int begin = interval[0].beginning;int end = interval[0].ending;for (int i = 1; i < n; i++){if (interval[i].beginning > begin && interval[i].beginning <= end){if (interval[i].ending <= end)continue;else end = interval[i].ending;}if (interval[i].beginning > end){cout << begin << " " << end << endl;begin = interval[i].beginning;end = interval[i].ending;}}cout << begin << " " << end;return 0;return 0;}