贪心+优先队列 HDOJ 5360 Hiking

题目传送门

 /*
     题意：求邀请顺序使得去爬山的人最多，每个人有去的条件
     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。
     详细解释：http://blog.csdn.net/queuelovestack/article/details/47319361
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-6 14:51:47
 * File Name     :H.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 struct Node {
     int l, r, id;
     bool operator < (const Node &x) const {
         return r > x.r;
     }
 }node[MAXN];
 priority_queue<Node> Q;
 bool vis[MAXN];
 int ans[MAXN];
 int n;

 bool cmp(Node x, Node y)    {
     if (x.l == y.l) return x.r < y.r;
     return x.l < y.l;
 }

 int main(void)    {     //HDOJ 5360 Hiking
     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%d", &n);
         for (int i=; i<=n; ++i)    {
             scanf ("%d", &node[i].l);   node[i].id = i;
         }
         for (int i=; i<=n; ++i)    {
             scanf ("%d", &node[i].r);
         }
         sort (node+, node++n, cmp);

         if (node[].l != ) {
             puts ("");
             for (int i=; i<=n; ++i)    printf ("%d%c", i, (i==n) ? '\n' : ' ');
             continue;
         }
         memset (vis, false, sizeof (vis));
         while (!Q.empty ()) Q.pop ();
         int i = , p = ;
         for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);
         while (!Q.empty ()) {
             while (!Q.empty ()) {
                 if (p <= Q.top ().r)    {
                     ans[++p] = Q.top ().id;
                     vis[ans[p]] = true;    Q.pop ();   break;
                 }
                 else Q.pop ();
             }
             for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);
         }

         printf ("%d\n", p);
         printf ("%d", ans[]);
         for (int i=; i<=p; ++i)    {
             printf (" %d", ans[i]);
         }
         for (int i=; i<=n; ++i)    {
             if (!vis[i])    printf (" %d", i);
         }
         puts ("");
     }

     return ;
 }