题目传送门
1 /*
2 题意：最少需要多少条线段能覆盖[0, m]的长度
3 贪心：首先忽略被其他线段完全覆盖的线段，因为选取更长的更优
4 接着就是从p=0开始，以p点为标志，选取 (node[i].l <= p && p < node[i+1].l)
5 详细解释：http://www.cnblogs.com/freezhan/p/3219046.html
6 */
7 #include <cstdio>
8 #include <iostream>
9 #include <algorithm>
10 #include <cstring>
11 #include <cmath>
12 using namespace std;
13
14 const int MAXN = 1e6 + 10;
15 const int INF = 0x3f3f3f3f;
16 struct Node
17 {
18 int l, r;
19 }node[MAXN], ans[MAXN];
20
21 bool cmp(Node x, Node y)
22 {
23 if (x.l == y.l) return x.r > y.r;
24 else return x.l < y.l;
25 }
26
27 int main(void) //URAL 1303 Minimal Coverage
28 {
29 //freopen ("R.in", "r", stdin);
30
31 int m;
32 while (scanf ("%d", &m) == 1)
33 {
34 int n = 0; int u, v;
35 while (scanf ("%d%d", &u, &v) == 2 && (u || v))
36 node[++n].l = u, node[n].r = v;
37 sort (node+1, node+1+n, cmp);
38
39 int k = 1;
40 for (int i=2; i<=n; ++i) //cover
41 {
42 if (node[k].l < node[i].l && node[k].r < node[i].r)
43 {
44 node[++k] = node[i];
45 }
46 }
47
48 n = k; int p = 0; int cnt = 0;
49 node[n+1].l = node[n+1].r = m + 1;
50 for (int i=1; i<=n; ++i)
51 {
52 if (node[i].l <= p && p < node[i+1].l)
53 {
54 ans[++cnt] = node[i]; p = node[i].r;
55 }
56 if (p >= m)
57 {
58 printf ("%d\n", cnt);
59 for (int i=1; i<=cnt; ++i)
60 {
61 printf ("%d %d\n", ans[i].l, ans[i].r);
62 }
63 break;
64 }
65 }
66
67 if (p < m) puts ("No solution");
68 }
69
70 return 0;
71 }
72
73 /*
74 No solution
75 */