题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/D
题目：
Given several segments of line (int the X axis) with coordinates [Li; Ri]. You are to choose the minimal
amount of them, such they would completely cover the segment [0; M].
Input
The rst 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 Li Ri"
(jLi
j; jRi
j 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.
Output
For each test case, in the rst 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 (Li), 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：
数轴上有n个闭区间[ai,bi]，选择尽量少的区间覆盖一条指定线段[0，m]。
分析：
贪心法
以0为起点，寻找包含在[0,m]间的最大区间的区间并且把[0,m]区间外的部分切掉，将没有修改前的区间保存到另一个数组中。
再以寻找出的区间的右端点为起点，进行查找。如此循环直到右端点大于等于m时停止。
注意个变量定义的区域
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include <algorithm> 5 using namespace std; 6 #define maxn 100010 7 struct node 8 { 9 int l; 10 int r; 11 }a[maxn],b[maxn]; 12 int cmp(const node &a, const node &b) //按l从小到大排序，当l相同时按r从大到小排序 13 { 14 if(a.l != b.l) return a.l < b.l; 15 else return a.r > b.r; 16 } 17 int j,m; 18 int main() 19 { 20 int t; 21 cin>>t; 22 while(t--) 23 { 24 memset(a,0,sizeof(a)); 25 memset(b,0,sizeof(b)); 26 int i=0; 27 int x,y; 28 cin>>m; 29 while(scanf("%d%d",&x,&y)) 30 { 31 if(!x&&!y) break; 32 a[i].l=x; 33 a[i].r=y; 34 i++; 35 } 36 j=i; 37 sort(a,a+j,cmp); 38 int f,pos; 39 int ma=0,k=0,l=0; 40 while(1) 41 { 42 if(l>=m) break; 43 f=0; 44 ma=0; 45 for(i=0;i<j;i++) 46 { 47 if(a[i].l<=l) 48 { 49 if(a[i].r>ma) 50 { 51 pos=i; //标记包含区间的位置 52 ma=a[i].r; //更改起点 53 f=1; 54 } 55 } 56 } 57 if(f) 58 { 59 l=ma; 60 b[k]=a[pos]; 61 k++; 62 } 63 else break; 64 } 65 if(f) 66 { 67 cout<<k<<endl; 68 for(i=0;i<k;i++) 69 cout<<b[i].l<<' '<<b[i].r<<endl; 70 if(t!=0) cout<<endl; 71 72 } 73 else { 74 cout<<'0'<<endl; 75 if(t!=0) cout<<endl;} 76 } 77 return 0; 78 }