bzoj2658 [Zjoi2012]小蓝的好友(mrx)

Posted 2020-06-17

题目链接

坑坑坑坑

我自己怎么想都不会把一列当成一个点写平衡树QAQ

扫描线+可持久化treap

showson教了我可持久化treap%%%

  1 #include<algorithm>
  2 #include<iostream>
  3 #include<cstdlib>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<string>
  7 #include<cmath>
  8 #include<ctime>
  9 #include<queue>
 10 #include<stack>
 11 #include<map>
 12 #include<set>
 13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
 14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
 15 #define Clear(a,b) memset(a,b,sizeof(a))
 16 #define inout(x) printf("%d",(x))
 17 #define douin(x) scanf("%lf",&x)
 18 #define strin(x) scanf("%s",(x))
 19 #define LLin(x) scanf("%lld",&x)
 20 #define op operator
 21 #define CSC main
 22 typedef unsigned long long ULL;
 23 typedef const int cint;
 24 typedef long long LL;
 25 using namespace std;
 26 void inin(int &ret)
 27 {
 28     ret=0;int f=0;char ch=getchar();
 29     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=getchar();}
 30     while(ch>=‘0‘&&ch<=‘9‘)ret*=10,ret+=ch-‘0‘,ch=getchar();
 31     ret=f?-ret:ret;
 32 }
 33 LL S(LL x){return x*(x+1)>>1;}
 34 int root,s[40040],h[40040],tag[40040],ch[40040][2],ed;
 35 LL ans[40040];
 36 int newnode(int x)
 37 {
 38     s[++ed]=1,h[ed]=x,ans[ed]=tag[ed]=0;
 39     return ed;
 40 }
 41 void maintain(int k)
 42 {
 43     if(!k)return ;
 44     s[k]=s[ch[k][0]]+s[ch[k][1]]+1;
 45     ans[k]=0;
 46     re(i,0,1)ans[k]+=ans[ch[k][i]]+S(s[ch[k][i]])*(h[ch[k][i]]-h[k]);
 47 }
 48 void add(int k,int d)
 49 {
 50     if(!k)return ;
 51     h[k]+=d,tag[k]+=d;
 52 }
 53 void down(int k)
 54 {
 55     add(ch[k][0],tag[k]);
 56     add(ch[k][1],tag[k]);
 57     tag[k]=0;
 58 }
 59 int merge(int l,int r)
 60 {
 61     if(!l||!r)return l+r;
 62     if(h[l]<h[r])
 63     {
 64         down(l);
 65         ch[l][1]=merge(ch[l][1],r);
 66         maintain(l);
 67         return l;
 68     }
 69     down(r);
 70     ch[r][0]=merge(l,ch[r][0]);
 71     maintain(r);
 72     return r;
 73 }
 74 pair<int,int> split(int k,int l)
 75 {
 76     if(!k)return pair<int,int>(0,0);
 77     pair<int,int> wocao;
 78     down(k);
 79     if(s[ch[k][0]]>=l)
 80     {
 81         wocao=split(ch[k][0],l);
 82         ch[k][0]=wocao.second;
 83         wocao.second=k;
 84     }
 85     else 
 86     {
 87         wocao=split(ch[k][1],l-s[ch[k][0]]-1);
 88         ch[k][1]=wocao.first;
 89         wocao.first=k; 
 90     }
 91     maintain(k);
 92     return wocao;
 93 }
 94 pair<int,int> p[100010];
 95 int main()
 96 {
 97     int r,c,n;
 98     inin(r),inin(c),inin(n);
 99     re(i,1,n)inin(p[i].first),inin(p[i].second);
100     sort(p+1,p+n+1);
101     re(i,1,c)root=merge(root,newnode(0));
102     LL an=S(r)*S(c);
103     int j=1;
104     re(i,1,r)
105     {
106         add(root,1);
107         while(j<=n&&p[j].first==i)
108         {
109             int x=p[j++].second;
110             pair<int,int> r1=split(root,x-1),r2=split(r1.second,1);
111             h[r2.first]=0;
112             root=merge(merge(r1.first,r2.first),r2.second);
113         }
114         an-=ans[root]+S(s[root])*h[root];
115     }
116     printf("%lld",an);
117      return 0;
118 }