贪心 UVALive 6834 Shopping

 

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\n", ans);
56     }
57 
58     return 0;
59 }

 

posted @ 2015-06-07 18:35  Running_Time  阅读(259)  评论(0)  编辑  收藏  举报