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线段树合并 || 树状数组 || 离散化 || BZOJ 4756: [Usaco2017 Jan]Promotion Counting || Luogu P3605 [USACO17JAN]Promotion Counting晋升者计数

题面:P3605 [USACO17JAN]Promotion Counting晋升者计数

题解:这是一道万能题,树状数组 || 主席树 || 线段树合并 || 莫队套分块 || 线段树 都可以写。。记得离散化

线段树合并版:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=(1e5)+50;
 7 int N,num_edge=0,edge_head[maxn],W[maxn],lsh_cnt=0,num_treenode=0;
 8 int root[maxn],ans[maxn],u;
 9 struct A_{int id,data;}A[maxn];
10 inline bool cmp(const A_&a,const A_&b){return a.data<b.data;}
11 struct Edge{int to,nx;}edge[maxn];
12 inline void Add_edge(int from,int to){
13     edge[++num_edge].nx=edge_head[from];
14     edge[num_edge].to=to;
15     edge_head[from]=num_edge;
16     return;
17 }
18 struct Tree{int lc,rc,l,r,cnt;}t[maxn*20];
19 inline void Build(int x,int l,int r,int q){
20     t[x].l=l;t[x].r=r;int mid=(l+r)>>1;
21     if(l==r&&l==q){
22         t[x].cnt=1;
23         return;
24     }
25     if(q<=mid)Build(t[x].lc=++num_treenode,l,mid,q);
26     else Build(t[x].rc=++num_treenode,mid+1,r,q);
27     t[x].cnt=t[t[x].lc].cnt+t[t[x].rc].cnt;
28     return;
29 }
30 inline int Merge(int u,int v){
31     if(!u)return v;
32     if(!v)return u;
33     int l=t[u].l,r=t[u].r;
34     if(l==r){
35         t[u].cnt+=t[v].cnt;
36         return u;
37     }
38     t[u].lc=Merge(t[u].lc,t[v].lc);
39     t[u].rc=Merge(t[u].rc,t[v].rc);
40     t[u].cnt=t[t[u].lc].cnt+t[t[u].rc].cnt;
41     return u;
42 }
43 inline void Query(int g,int x,int ql,int qr){
44     int l=t[x].l,r=t[x].r,mid=(l+r)>>1,lc=t[x].lc,rc=t[x].rc;
45     if(x==0)return;
46     if(ql<=l&&r<=qr){
47         ans[g]+=t[x].cnt;
48         return;
49     }
50     if(ql<=mid)Query(g,lc,ql,qr);
51     if(qr>mid) Query(g,rc,ql,qr);
52     return;
53 }
54 inline void Dfs(int x){
55     for(int i=edge_head[x];i;i=edge[i].nx){
56         int y=edge[i].to;
57         Dfs(y);
58         root[x]=Merge(root[x],root[y]);
59     }
60     if(W[x]+1>lsh_cnt)ans[x]=0;
61     else Query(x,root[x],W[x]+1,lsh_cnt);
62     return;
63 }
64 int main(){
65     scanf("%d",&N);
66     for(int i=1;i<=N;i++){
67         scanf("%d",&A[i].data);
68         A[i].id=i;
69     }
70     sort(A+1,A+N+1,cmp);
71     W[A[1].id]=++lsh_cnt;
72     for(int i=2;i<=N;i++)
73         if(A[i].data!=A[i-1].data)W[A[i].id]=++lsh_cnt;
74         else W[A[i].id]=lsh_cnt;
75     for(int i=2;i<=N;i++){
76         scanf("%d",&u);
77         Add_edge(u,i);
78     }
79     for(int i=1;i<=N;i++)Build(root[i]=++num_treenode,1,lsh_cnt+5,W[i]);
80     Dfs(1);
81     for(int i=1;i<=N;i++)printf("%d\n",ans[i]);
82     return 0;
83 }

树状数组版:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #define ll long long
 6 #define re register
 7 using namespace std;
 8 inline int rd(){
 9     int x=0;char c=getchar();
10     while(c<'0'||c>'9')c=getchar();
11     while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
12     return x;
13 }
14 const int maxn=1e5;
15 int N,P[maxn+5],cnt=0,num_edge=0,edge_head[maxn+5],F,C[maxn+5];
16 ll ans[maxn+5];
17 struct Edge{
18     int to,nx;
19 }edge[maxn+5];
20 struct Node{
21     int x,id;
22 }A[maxn+50];
23 inline bool cmp(const Node&a,const Node&b){
24     if(a.x<b.x)return 1;
25     return 0;
26 }
27 inline int Find(int x){
28     re ll ans=0;
29     for(;x<=cnt;x+=x&(-x))ans+=C[x];
30     return ans;
31 }
32 inline void Update(int x){
33     for(;x>0;x-=x&(-x))C[x]++;
34     return;
35 }
36 inline void Dfs(int x){
37     ans[x]-=Find(P[x]);
38     for(re int i=edge_head[x];i;i=edge[i].nx) Dfs(edge[i].to);     
39     ans[x]+=Find(P[x]);
40     Update(P[x]);
41     return;
42 }
43 int main(){
44     N=rd();
45     for(re int i=1;i<=N;i++){
46         A[i].x=rd();
47         A[i].id=i;
48     }
49     sort(A+1,A+N+1,cmp);
50     for(re int i=1;i<=N;i++){
51         if(i==1||A[i].x!=A[i-1].x)cnt++;
52         P[A[i].id]=cnt;
53     }
54     for(re int i=2;i<=N;i++){
55         scanf("%d",&F);
56         edge[++num_edge].nx=edge_head[F];
57            edge[num_edge].to=i;
58         edge_head[F]=num_edge;
59     }
60     Dfs(1);
61     for(re int i=1;i<=N;i++)printf("%lld\n",ans[i]);
62     return 0;
63 }

By:AlenaNuna

posted on 2019-03-13 20:09  AlenaNuna  阅读(226)  评论(0编辑  收藏  举报