贪心+优先队列 HDOJ 5360 Hiking

 1 /*
 2     题意：求邀请顺序使得去爬山的人最多，每个人有去的条件
 3     贪心+优先队列：首先按照l和r从小到大排序，每一次将当前人数相同的被邀请者入队，那么只要能当前人数比最多人数条件小，该人能                 被邀请，而且不用考虑最少人数的条件。网上的写的比我简洁，学习了。
 4     详细解释：http://blog.csdn.net/queuelovestack/article/details/47319361
 5 */
 6 /************************************************
 7 * Author        :Running_Time
 8 * Created Time  :2015-8-6 14:51:47
 9 * File Name     :H.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e5 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 struct Node {
38     int l, r, id;
39     bool operator < (const Node &x) const {
40         return r > x.r;
41     }
42 }node[MAXN];
43 priority_queue<Node> Q;
44 bool vis[MAXN];
45 int ans[MAXN];
46 int n;
47 
48 bool cmp(Node x, Node y)    {
49     if (x.l == y.l) return x.r < y.r;
50     return x.l < y.l;
51 }
52 
53 int main(void)    {     //HDOJ 5360 Hiking
54     int T;  scanf ("%d", &T);
55     while (T--) {
56         scanf ("%d", &n);
57         for (int i=1; i<=n; ++i)    {
58             scanf ("%d", &node[i].l);   node[i].id = i;
59         }
60         for (int i=1; i<=n; ++i)    {
61             scanf ("%d", &node[i].r);
62         }
63         sort (node+1, node+1+n, cmp);
64 
65         if (node[1].l != 0) {
66             puts ("0");
67             for (int i=1; i<=n; ++i)    printf ("%d%c", i, (i==n) ? '
' : ' ');
68             continue;
69         }
70         memset (vis, false, sizeof (vis));
71         while (!Q.empty ()) Q.pop ();
72         int i = 1, p = 0;
73         for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);
74         while (!Q.empty ()) {
75             while (!Q.empty ()) {
76                 if (p <= Q.top ().r)    {
77                     ans[++p] = Q.top ().id; 
78                     vis[ans[p]] = true;    Q.pop ();   break;
79                 }
80                 else Q.pop ();
81             }
82             for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);
83         }
84         
85         printf ("%d
", p);
86         printf ("%d", ans[1]);
87         for (int i=2; i<=p; ++i)    {
88             printf (" %d", ans[i]);
89         }
90         for (int i=1; i<=n; ++i)    {
91             if (!vis[i])    printf (" %d", i);
92         }
93         puts ("");
94     }
95 
96     return 0;
97 }
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