LOJ P3960 列队 树状数组 vector

https://www.luogu.org/problemnew/show/P3960

树状数组预处理之后直接搞就可以了,也不是很好解释,反正就是一个模拟过程的暴力用树状数组维护,还挺巧妙的。

我为什么考场上想不出来嘤嘤嘤

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<vector>
 7 using namespace std;
 8 const int maxn=600010;
 9 int n,m,q;
10 int xx[maxn]={},yy[maxn]={};
11 int num[maxn]={},z[maxn]={};
12 int lag[maxn]={};
13 int t[maxn]={},mx;
14 vector<int>id[maxn];
15 vector<int>val[maxn];
16 vector<long long >shu[maxn];
17 void add(int x,int v){
18     while(x<=mx){
19         t[x]+=v;
20         x+=(x&-x);
21     }
22 }
23 int getit(int x){
24     int tsn=0;
25     while(x){
26         tsn+=t[x];
27         x-=(x&-x);
28     }return tsn;
29 }
30 int Find(int x){
31     int l=0,r=mx,mid;
32     while(l<r){
33         mid=(l+r)/2;
34         if(getit(mid)<x)l=mid+1;
35         else r=mid;
36     }
37     return l;
38 }
39 int main(){
40     scanf("%d%d%d",&n,&m,&q);
41     mx=max(n,m)+q;
42     for(int i=1;i<=q;i++){
43         scanf("%d%d",&xx[i],&yy[i]);
44         if(yy[i]!=m){
45             val[xx[i]].push_back(yy[i]);
46             id[xx[i]].push_back(i);
47         }
48         else
49             z[i]=1;
50     }
51     for(int i=1;i<=mx;i++)add(i,1);
52     for(int i=1;i<=n;i++){
53         int siz=id[i].size();
54         for(int j=0;j<siz;j++){
55             num[id[i][j]]=Find(val[i][j]);
56             add(num[id[i][j]],-1);
57         }
58         for(int j=0;j<siz;j++)add(num[id[i][j]],1);
59     }
60     long long ans;
61     for(int i=0;i<=n;i++)shu[0].push_back((long long)i*m);
62     lag[0]=m;
63     for(int i=1;i<=q;i++){
64         if(z[i]){
65             ans=shu[0][Find(xx[i])];
66             add(Find(xx[i]),-1);
67         }
68         else if(num[i]<=m+lag[xx[i]]-1){
69             ans=num[i]<m?(long long)(xx[i]-1)*m+num[i]:shu[xx[i]][num[i]-m];
70             shu[xx[i]].push_back(shu[0][Find(xx[i])]);
71             lag[xx[i]]++;
72             add(Find(xx[i]),-1);
73         }
74         else{
75             ans=shu[0][Find(num[i]-m-lag[xx[i]]+xx[i])];
76             add(Find(num[i]-m-lag[xx[i]]+xx[i]),-1);
77         }
78         shu[0].push_back(ans);lag[0]++;
79         printf("%lld
",ans);
80     }
81     return 0;
82 }
View Code

原文地址:https://www.cnblogs.com/137shoebills/p/8674514.html