【bzoj 3262】陌上花开

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陌上花开

Time Limit: 20 Sec Memory Limit: 256 MB
Submit: 2074 Solved: 921
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Description

有n朵花，每朵花有三个属性：花形(s)、颜色(c)、气味(m)，又三个整数表示。现要对每朵花评级，一朵花的级别是它拥有的美丽能超过的花的数量。定义一朵花A比另一朵花B要美丽，当且仅当Sa>=Sb,Ca>=Cb,Ma>=Mb。显然，两朵花可能有同样的属性。需要统计出评出每个等级的花的数量。
Input

第一行为N,K (1 <= N <= 100,000, 1 <= K <= 200,000 ), 分别表示花的数量和最大属性值。
以下N行，每行三个整数si, ci, mi (1 <= si, ci, mi <= K)，表示第i朵花的属性
Output

包含N行，分别表示评级为0…N-1的每级花的数量。
Sample Input

10 3

3 3 3

2 3 3

2 3 1

3 1 1

3 1 2

1 3 1

1 1 2

1 2 2

1 3 2

1 2 1

Sample Output

HINT

1 <= N <= 100,000, 1 <= K <= 200,000

Source

树套树 CDQ分治
作死写树套树，好吧我输了，以后补一补cdq分治

//unsolve#include<bits/stdc++.h>#define __de__using namespace std;const int maxn = 200000;int pool[maxn],cnt;int lch[maxn],rch[maxn],w[maxn],v[maxn],s[maxn],rnd[maxn];int cntt[maxn];int N,M;struct data{    int a,b,c,id;   };data in[maxn];bool cmp(data _a,data _b){    if(_a.a==_b.a)    {        if(_a.b==_b.b)        {            return _a.c<_b.c;        }        else        {            return _a.b<_b.b;        }    }    return _a.a<_b.a;}struct treap{    int root;    void pushup(int nd)    {        s[nd]=s[lch[nd]]+s[rch[nd]]+v[nd];    }    void rturn(int &nd)    {        int k=lch[nd];        lch[nd]=rch[k];        rch[k]=nd;        s[k]=s[nd];//      pushup(nd);        pushup(nd);        nd=k;       }    void lturn(int &nd)    {        int k=rch[nd];        rch[nd]=lch[k];        lch[k]=nd;    //  pushup(nd);        s[k]=s[nd];        pushup(nd);        nd=k;    }    void insert(int wc)    {        insert(this->root,wc);    }    int getrank(int wc)    {        return getrank(this->root,wc);      }    void insert(int &nd,int wc)    {        if(nd==0)        {            nd=++cnt;            w[nd]=wc;            s[nd]++;v[nd]++;            rnd[nd]=rand();            return ;                    }        s[nd]++;        if(w[nd]==wc)        {            v[nd]++;            return;        }        else if(wc<w[nd])        {            insert(lch[nd],wc);            if(rnd[lch[nd]] > rnd[nd])            {                rturn(nd);            }        }        else        {            insert(rch[nd],wc);            if(rnd[rch[nd]] > rnd[nd])            {                lturn(nd);            }        }    }    int getrank(int nd,int wc)    {        if(nd==0)return -0x7fffffff;        if(w[nd]==wc)return v[nd]+s[lch[nd]];        else if(w[nd]>wc)        {            return getrank(lch[nd],wc);        }        else            return s[lch[nd]]+v[nd]+getrank(rch[nd],wc);    }    void dfs()    {        dfs(this->root);    }    void dfs(int nd)    {        if(nd==0)return;        dfs(lch[nd]);cout<<rch[nd]<<" "<<lch[nd]<<" "<<s[nd]<<" "<<w[nd]<<" "<<v[nd]<<" "<<nd<<endl;        dfs(rch[nd]);    }};treap fenwick[maxn];void modify(int n,int num){    while(n<=M)    {        cout<<"&&&&&&&&&&&&&&&\n"<<n<<"\n";        fenwick[n].dfs();        fenwick[n].insert(num);        cout<<"after insert\n";        fenwick[n].dfs();        n+=(n&-n);    }}int sum(int n,int num){    int ans=0;    while(n>0)    {        ans+=fenwick[n].getrank(num);        n-=(n&-n);    }}#ifndef __de__int main(){    fenwick[0].insert(76);    fenwick[0].insert(76);    fenwick[0].dfs();    cout<<fenwick[0].getrank(76);    return 0;}#endif#ifdef __de__int main(){    scanf("%d%d",&N,&M);    for(int i=1;i<=N;i++)    {        scanf("%d%d%d",&in[i].a,&in[i].b,&in[i].c);    }    sort(in+1,in+N+1,cmp);    for(int i=1;i<=N;i++)    {        cout<<in[i].b<<" "<<in[i].c<<endl;        if(in[i].a==in[i+1].a&&in[i].b==in[i+1].b&&in[i].c==in[i+1].c&&i!=N)        {            cntt[i+1]+=cntt[i]+1;        }        else        {            int rk=sum(in[i].b,in[i].c);            cout<<rk<<endl;            cntt[rk]+=cntt[i]+1;        }        modify(in[i].b,in[i].c);    }    for(int i=1;i<=N;i++)        //cout<<cntt[i]<<endl;    return 0;}#endif

对着别人写的抄了一遍

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct flower{    int x,y,z;    int cnt,ans;}d[1000010];int tree[3000010],n,k,tot,num[1000010];int tmp(flower a,flower b){    if(a.x<b.x) return 1;    if(a.x>b.x) return 0;    if(a.y<b.y) return 1;    if(a.y>b.y) return 0;    if(a.z<b.z) return 1;    return 0;}int cmp(flower a,flower b){    if(a.y<b.y) return 1;    if(a.y>b.y) return 0;    if(a.z<b.z) return 1;    if(a.z>b.z) return 0;    if(a.x<b.x) return 1;    return 0;}int lowbit(int x){    return x&(-x);}void updata(int x,int v){    while(x<=k)     {        tree[x]+=v;        x+=lowbit(x);     }    return;}inline int ask(int x){    int sum=0;    while(x)     {        sum+=tree[x];        x-=lowbit(x);     }    return sum;}void CDQ(int l,int r){    if(l==r)      {        d[l].ans+=d[l].cnt-1;        return;     }    int mid=(l+r)>>1;    CDQ(l,mid); CDQ(mid+1,r);    sort(d+l,d+mid+1,cmp);    sort(d+mid+1,d+r+1,cmp);    int j=l;    for(int i=mid+1;i<=r;++i)     {        while(j<=mid&&d[j].y<=d[i].y)           updata(d[j].z,d[j].cnt),++j;         d[i].ans+=ask(d[i].z);     }    for(int i=l;i<j;++i) updata(d[i].z,-d[i].cnt);}int main(){    int i,j;    scanf("%d%d",&n,&k);    for(i=1;i<=n;++i) scanf("%d%d%d",&d[i].x,&d[i].y,&d[i].z),d[i].ans=1;    sort(d+1,d+n+1,tmp);    for(i=1;i<=n;++i)     if(i!=1&&d[i].x==d[i-1].x&&d[i].y==d[i-1].y&&d[i].z==d[i-1].z) d[tot].cnt++;      else d[++tot]=d[i],d[tot].cnt=1;    CDQ(1,tot);    sort(d+1,d+tot+1,tmp);    for(i=1;i<=tot;++i) num[d[i].ans]+=d[i].cnt;    for(i=1;i<=n;++i) printf("%d\n",num[i]);    return 0;}

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