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  • HDU 5360 Hiking (贪心)

    题意:邀请 n 参加聚会,如果在邀请第 i 个人之前,已经成功邀请了 x 个人,并且 li <= x <= ri,那么第 i 人才会去,问你怎么排列使得邀请的人最多。

    析:对于所有的人,按照 li 进行排序,对于维护一个优先队列,队列内是 ri 小的优先,然后枚举每个时间点,把 li 等于的当前时间点的都放到队列中,然后取出最优先那个,这就是要邀请的,当然如果队列顶上的 ri 小于当前时间点,要把它从队列中删除。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #include <numeric>
    #define debug() puts("++++")
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define be begin()
    #define ed end()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define lowbit(x) -x&x
    //#define all 1,n,1
    #define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
    #define freopenr freopen("in.in", "r", stdin)
    #define freopenw freopen("out.out", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const LL LNF = 1e17;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e5 + 20;
    const int maxm = 1e6 + 10;
    const LL mod = 1000000000000000LL;
    const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
    const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c) {
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    inline int readInt(){ int x;  scanf("%d", &x);  return x; }
    
    struct Node{
      int l, r, id;
      bool operator < (const Node &rhs) const{
        return r > rhs.r;
      }
    };
    
    Node a[maxn];
    bool vis[maxn];
    
    
    int main(){
      int T;  cin >> T;
      while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
          scanf("%d", &a[i].l);
          a[i].id = i;
        }
        for(int i = 0; i < n; ++i)  scanf("%d", &a[i].r);
        sort(a, a + n, [&](Node a, Node b){ return a.l < b.l; });
        priority_queue<Node> pq;
        vector<int> ans;
        int k = 0;
        for(int i = 0; i < n; ++i){
          while(ans.sz == i && k < n && a[k].l == i)  pq.push(a[k++]);
          while(!pq.empty() && pq.top().r < i)  pq.pop();
          if(pq.empty())  continue;
          ans.pb(pq.top().id);  pq.pop();
        }
        printf("%d
    ", ans.sz);
        int cnt = 0;  ms(vis, 0);
        for(int i = 0; i < ans.sz; ++i){
          vis[ans[i]] = 1;
          if(cnt)  printf(" %d", ans[i] + 1);
          else printf("%d", ans[i] + 1);
          cnt = 1;
        }
        for(int i = 0; i < n; ++i)  if(!vis[i]){
          if(cnt)  printf(" %d", i + 1);
          else printf("%d", i + 1);
          cnt = 1;
        }
        printf("
    ");
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/8974144.html
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