题目传送门
1 /*
2 题意：有n个商店排成一条直线，有一些商店有先后顺序，问从0出发走到n+1最少的步数
3 贪心：对于区间被覆盖的点只进行一次计算，还有那些要往回走的区间步数*2，再加上原来最少要走n+1步就是答案了
4 详细解释：http://blog.csdn.net/u013625492/article/details/45640735
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cstring>
9 #include <cmath>
10 using namespace std;
11
12 const int MAXN = 1e3 + 10;
13 const int INF = 0x3f3f3f3f;
14 struct A
15 {
16 int l, r;
17 }a[MAXN];
18 struct B
19 {
20 int l, r;
21 }b[MAXN];
22
23 bool cmp(A x, A y) {return x.l < y.l;}
24
25 int main(void) //UVALive 6834 Shopping
26 {
27 // freopen ("C.in", "r", stdin);
28
29 int n, m;
30 while (scanf ("%d%d", &n, &m) == 2)
31 {
32 int cnt = 0;
33 for (int i=1; i<=m; ++i)
34 {
35 int u, v; scanf ("%d%d", &u, &v);
36 if (u > v) continue;
37 a[++cnt].l = u; a[cnt].r = v;
38 }
39 sort (a+1, a+1+cnt, cmp);
40
41 int l = a[1].l; int r = a[1].r; int tot = 0;
42 for (int i=2; i<=cnt; ++i) //区间覆盖
43 {
44 if (r < a[i].l)
45 {
46 b[++tot].l = l; b[tot].r = r;
47 l = a[i].l; r = a[i].r;
48 }
49 else r = max (r, a[i].r);
50 if (i == cnt) {b[++tot].l = l; b[tot].r = r;}
51 }
52
53 int ans = n + 1;
54 for (int i=1; i<=tot; ++i) ans += 2 * (b[i].r - b[i].l);
55 printf ("%d
", ans);
56 }
57
58 return 0;
59 }