Minimal coverage 

The Problem

Given several segments of line (int the X axis) with coordinates [L_i,R_i]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M].

The Input

The first line is the number of test cases, followed by a blank line.

Each test case in the input should contains an integer M(1<=M<=5000), followed by pairs "L_i R_i"(|L_i|, |R_i|<=50000, i<=100000), each on a separate line. Each test case of input is terminated by pair "0 0".

Each test case will be separated by a single line.

The Output

For each test case, in the first line of output your programm should print the minimal number of line segments which can cover segment [0,M]. In the following lines, the coordinates of segments, sorted by their left end (L_i), should be printed in the same format as in the input. Pair "0 0" should not be printed. If [0,M] can not be covered by given line segments, your programm should print "0"(without quotes).

Print a blank line between the outputs for two consecutive test cases.

Sample Input

2

1
-1 0
-5 -3
2 5
0 0

1
-1 0
0 1
0 0

Sample Output

0

1
0 1

题意：用最少的区间覆盖（0，m）这个区间。

分析：尽量选覆盖目的区间大的区间。我们可以按照每个给的区间的最左端端点排序，从左端点小于st的区间中选取右端点最大的区间赋给en。这时候再比较st=en与m的大小，不满足k>=m，继续上面的循环，直到满足，或者找不到可以覆盖的区间。 详情看代码。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define M 100005
using namespace std;
struct node{
	int st, en;
}s[M];
int ans[M];

int cmp(node a, node b){
	if(a.st == b.st) return a.en>b.en;
	return a.st < b.st;
}
int main(){
	int t, m;
	scanf("%d", &t);
	while(t --){
		scanf("%d", &m);
		int tot = 0, a, b;
		while(scanf("%d%d", &a, &b), a||b){
			if(a>b) swap(a, b);
			s[tot].st = a; s[tot].en = b;
			++tot;
		}
		sort(s, s+tot, cmp);
		int st, en, num;
		st = en = num = 0;
		while(st<m){
			en = st;
			for(int i = 0; i < tot; i ++){
				if(s[i].st<=st&&s[i].en>en){  //从左端点小于st的区间中找出右端点最大的赋给en，并且用ans【num】储存右端点最大的区间的下标
					en = s[i].en; ans[num] = i; 
				}
			}
			if(en == st){  //如果没有找到，可以继续覆盖的区间，直接跳出来。
				num = 0; break;
			}
			st = en;
			++num;
		}
		cout<<num<<endl;
		for(int i = 0; i < num; i++)
			cout<<s[ans[i]].st<<" "<<s[ans[i]].en<<endl;
	}
	exit(0);
}

uva 10020 Minimal coverage 【贪心】+【区间完全覆盖】