BZOJ 5110: [CodePlus2017]Yazid 的新生舞会 线段树

非常好的一道思维题.   

code:

#include <cstdio> 
#include <cstring> 
#include <algorithm> 
#include <vector>  
#define lson x<<1 
#define rson x<<1|1    
#define N 500010  
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)  
using namespace std;        
ll ans; 
int n; 
vector<int>p[N];   
int s1[N<<3],z1[N<<3];   
ll s2[N<<3],z2[N<<3];   
int tag[N<<3];   
bool clr[N<<3];    
void pushdown(int x) 
{
    if(clr[x]) 
    {
        clr[lson]=clr[rson]=1,s1[lson]=s1[rson]=s2[lson]=s2[rson]=tag[lson]=tag[rson]=clr[x]=0; 
    }
    if(tag[x]) 
    {    
        s1[lson]+=z1[lson]*tag[x],s1[rson]+=z1[rson]*tag[x]; 
        s2[lson]+=z2[lson]*tag[x],s2[rson]+=z2[rson]*tag[x];  
        tag[lson]+=tag[x],tag[rson]+=tag[x];  
        tag[x]=0; 
    }
}
void build(int l,int r,int x) 
{
    if(l==r) 
    {
        z1[x]=1,z2[x]=l; 
        return; 
    }
    int mid=(l+r)>>1;   
    build(l,mid,lson),build(mid+1,r,rson);   
    z1[x]=z1[lson]+z1[rson]; 
    z2[x]=z2[lson]+z2[rson];  
}
void update(int l,int r,int x,int L,int R) 
{
    if(l>=L&&r<=R) 
    {   
        s1[x]+=z1[x],s2[x]+=z2[x],++tag[x];  
        return; 
    } 
    pushdown(x); 
    int mid=(l+r)>>1;   
    if(L<=mid)    update(l,mid,lson,L,R); 
    if(R>mid)     update(mid+1,r,rson,L,R);   
    s1[x]=s1[lson]+s1[rson],s2[x]=s2[lson]+s2[rson];  
}
int query1(int l,int r,int x,int L,int R) 
{
    if(l>=L&&r<=R) return s1[x];  
    pushdown(x); 
    int mid=(l+r)>>1,re=0; 
    if(L<=mid)   re+=query1(l,mid,lson,L,R); 
    if(R>mid)    re+=query1(mid+1,r,rson,L,R); 
    return re; 
}
ll query2(int l,int r,int x,int L,int R) 
{
    if(l>=L&&r<=R) return s2[x];  
    pushdown(x); 
    int mid=(l+r)>>1;  
    ll re=0; 
    if(L<=mid)  re+=query2(l,mid,lson,L,R); 
    if(R>mid)   re+=query2(mid+1,r,rson,L,R); 
    return re; 
}
int main() 
{
    // setIO("input");    
    scanf("%d",&n);  
    int yy; scanf("%d",&yy); 
    int i,j,l,r,sum;  
    for(i=1;i<=n;++i) scanf("%d",&j), p[j].push_back(i);   
    build(-n,n,1);   
    for(i=0;i<n;++i) if(p[i].size()) 
    {
        s1[1]=s2[1]=tag[1]=sum=0,clr[1]=1;      
        p[i].push_back(n+1);    
        update(-n,n,1,0,0);    
        for(j=0;j<p[i].size();++j) 
        {
            if((!j&&p[i][j]>1)||(j&&p[i][j]>p[i][j-1]+1)) 
            {
                l=(!j)?1:(p[i][j-1]+1),r=p[i][j]-1;      
                if(j) ans+=query1(-n,n,1,-n,sum-1)*(r-l+1)-query2(-n,n,1,sum-(r-l+1),sum-1)+query1(-n,n,1,sum-(r-l+1),sum-1)*(sum-1-(r-l+1));   
                update(-n,n,1,sum-(r-l+1),sum-1);     
                sum-=(r-l+1);   
            }
            if(j!=p[i].size()-1) ++sum,ans+=query1(-n,n,1,-n,sum-1),update(-n,n,1,sum,sum);  
        }
    }
    printf("%lld\n",ans); 
    return 0; 
}