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; }