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  • 树形DP+树状数组 HDU 5877 Weak Pair

     1 //树形DP+树状数组 HDU 5877  Weak Pair
     2 // 思路:用树状数组每次加k/a[i],每个节点ans+=Sum(a[i]) 表示每次加大于等于a[i]的值
     3 // 这道题要离散化
     4 
     5 #include <bits/stdc++.h>
     6 using namespace std;
     7 #define LL long long
     8 typedef pair<int,int> pii;
     9 const double inf = 123456789012345.0;
    10 const LL MOD =100000000LL;
    11 const int N = 2e5+10;
    12 const int maxx = 200010; 
    13 #define clc(a,b) memset(a,b,sizeof(a))
    14 const double eps = 1e-7;
    15 void fre() {freopen("in.txt","r",stdin);}
    16 void freout() {freopen("out.txt","w",stdout);}
    17 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
    18 
    19 map<LL,LL> ma;
    20 LL a[N];
    21 LL c[N],b[N];
    22 LL in[N];
    23 vector<LL> g[N];
    24 LL lowbit(LL x){ return x&(-x);}
    25 LL add(LL x,int t){
    26     while(x>0){
    27        c[x]+=t;
    28        x-=lowbit(x);
    29     }
    30 }
    31 LL Sum(LL x){
    32     LL sum=0;
    33     while(x<maxx){
    34         sum+=c[x];
    35         x+=lowbit(x);
    36     }
    37     return sum;
    38 }
    39 
    40 LL ans=0;
    41 LL n,k;
    42 void dfs(LL rt){
    43      for(LL i=0;i<(int)g[rt].size();i++){
    44          LL v=g[rt][i];
    45          ans+=Sum(ma[a[v]]);
    46          if(a[v]==0) add(maxx,1);
    47          else add(ma[k/a[v]],1);
    48          dfs(v);
    49          if(a[v]==0) add(maxx,-1);
    50          else add(ma[k/a[v]],-1);
    51      }
    52 }
    53 int main(){
    54     int T;
    55     scanf("%d",&T);
    56     while(T--){
    57         ma.clear();
    58         memset(c,0,sizeof(c));
    59         scanf("%I64d%I64d",&n,&k);
    60         for(int i=1;i<=n;i++){
    61             scanf("%I64d",&a[i]);
    62             b[i*2-2]=a[i];
    63             if(a[i]!=0) b[i*2-1]=k/a[i];
    64             g[i].clear();
    65             in[i]=0;
    66         }
    67         sort(b,b+2*n);
    68         int K=unique(b,b+2*n)-b;
    69         int cxt=0;
    70         for(int i=0;i<K;i++){
    71             ma[b[i]]=++cxt;
    72         }
    73         for(LL i=0;i<n-1;i++){
    74             LL u,v;
    75             scanf("%I64d%I64d",&u,&v);
    76             g[u].push_back(v);
    77             in[v]++;
    78         }
    79         LL rt;
    80         for(LL i=1;i<=n;i++){
    81             if(in[i]==0){
    82                 rt=i;
    83                 break;
    84             }
    85         }
    86         ans=0;
    87         if(a[rt]==0) add(maxx,1);
    88         else add(ma[k/a[rt]],1);
    89         dfs(rt);
    90         printf("%I64d
    ",ans);
    91     }
    92     return 0;
    93 }
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  • 原文地址:https://www.cnblogs.com/ITUPC/p/5861453.html
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