P3960 列队

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #define ll long long
 6 using namespace std; 
 7 
 8 template <typename T> void in(T &x) {
 9     x = 0; T f = 1; char ch = getchar();
10     while(!isdigit(ch)) {if(ch == ‘-‘) f = -1; ch = getchar();}
11     while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
12     x *= f;
13 }
14 
15 template <typename T> void out(T x) {
16     if(x < 0) x = -x , putchar(‘-‘);
17     if(x > 9) out(x/10);
18     putchar(x%10 + 48);
19 }
20 //-------------------------------------------------------------
21 
22 const int N = 3e5+7;
23 
24 int n,m,q,x,y,now,maxn;
25 int end[N];
26 
27 struct node {
28     int lc,rc,sz;ll val;
29 }t[6000000];int cnt,rt[N];
30 
31 int get_sz(int l,int r) {
32     if(now == n + 1) {
33         if(r <= n) return r - l + 1;
34         if(l <= n) return n - l + 1;
35         return 0;
36     }
37     if(r <= m-1) return r - l + 1;
38     if(l <= m-1) return (m-1)-l+1;
39     return 0;
40 }
41 
42 ll get_val(int j) {
43     return (now == n+1) ? (1ll*j*m) : (1ll*(now-1)*m+j);
44 }
45 
46 ll D(int &u,int l,int r,int k) {
47     if(!u) {
48         u = ++cnt; t[u].sz = get_sz(l,r);
49         if(l == r) t[u].val = get_val(l);
50     }
51     --t[u].sz;
52     if(l == r) return t[u].val;
53     int mid = (l+r)>>1;
54     int l_sz = (t[u].lc ? t[t[u].lc].sz : get_sz(l,mid));
55     if(k <= l_sz) return D(t[u].lc,l,mid,k);
56     else return D(t[u].rc,mid+1,r,k-l_sz);
57 }
58 
59 void A(int &u,int l,int r,int pos,ll w) {
60     if(!u) {
61         u = ++cnt; 
62         t[u].sz = get_sz(l,r);
63     }
64     ++t[u].sz;//标记永久化的写法 
65     if(l == r) {
66         t[u].val = w; return;
67     }
68     int mid = (l+r)>>1;
69     if(pos <= mid) A(t[u].lc,l,mid,pos,w);
70     else A(t[u].rc,mid+1,r,pos,w);
71 }
72 
73 int main() {
74     freopen("0.in","r",stdin);
75     //freopen("my19.out","w",stdout);
76     ll id; in(n); in(m); in(q);//debug ll -> int
77     maxn = max(m,n)+q;//debug max(m,n) -> m!!!!!
78     int ans = 0;
79     while(q--) {
80         in(x); in(y);
81         cout << ++ans << ‘:‘ << ‘ ‘;
82         if(y == m) {
83             now = n+1; id = D(rt[n+1],1,maxn,x); out(id),putchar(‘\n‘);
84             A(rt[n+1],1,maxn,n+(++end[n+1]),id);
85         }
86         else {
87             now = x; id = D(rt[x],1,maxn,y); out(id),putchar(‘\n‘);
88             now = n+1; A(rt[n+1],1,maxn,n+(++end[n+1]),id);
89             now = n+1; id = D(rt[n+1],1,maxn,x);
90             now = x; A(rt[x],1,maxn,m-1+(++end[x]),id);
91         }
92     }
93     return 0;
94 }