Hdu 4605-Magic Ball Game (tree array)

Question:

Here is a binary tree. Each node has a w value. Now there is a small ball with the value of x, which is dropped from the root node. If w = x, it will stop; if w> x, the probability of its son going to the left and right is 1 and 2. If w <x, the probability of its son going to the left is 1/8, and that of its right son is 7/8. Now, let's ask you how much u is possible for a ball channel with the value of x.

Ideas:

Use a tree array for offline processing.

Excerpted by the senior student:

"There is a unique road from a node u to a point v. We only need the w value when it turns left on this road (X may be greater than the grtL value of this point, maybe less than the lessL value of this point) in the case of turning to the right, we can know the value of w (X may be greater than the point's weight grtR, or less than the point's weight lessR:

X = grtR (only contributing to 7) y = (lessL + lessR) + (grtL + grtR) * 3 ;"

 

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<iostream>#include<vector>#include<algorithm>#include<string.h>using namespace std;#define maxn 100005struct list{int l,r;int w;}node[maxn];int fs[maxn*2];int fss[maxn*2];struct qq{int x;int id;}xx;vector<qq>num[maxn];int ans[maxn][2];int sum[maxn][2];int w[maxn];int n,m,q,len;int lowbit(int x){ return (x&(-x));}int search(int l,int r,int w){while(l<r){int mid=(l+r)/2;if(fs[mid]==w)return mid;if(fs[mid]<w)l=mid+1;else r=mid;}}void add(int x,int bs,int num){for(;x<len;x+=lowbit(x)){sum[x][bs]+=num;}}int allsum(int s,int bs){int ss;ss=0;while(s>0){ss+=sum[s][bs];s=s-lowbit(s);}return ss;}void dfs(int x){int s;s=num[x].size();for(int i=0;i<s;i++){xx=num[x][i];int z,id;z=search(1,len,xx.x);id=xx.id;if(allsum(z,0)-allsum(z-1,0)+allsum(z,1)-allsum(z-1,1)>0){ans[id][0]=-1;continue;}int ll,lr,rl,rr;ll=allsum(len-1,0)-allsum(z,0);rl=allsum(z,0);lr=allsum(len-1,1)-allsum(z,1);rr=allsum(z,1);ans[id][0]=rr;ans[id][1]=(rl+rr)*3+ll+lr;}s=search(1,len,node[x].w);if(node[x].l!=-1){add(s,0,1);dfs(node[x].l);add(s,0,-1);}if(node[x].r!=-1){add(s,1,1);dfs(node[x].r);add(s,1,-1);}}int main(){int T,a,b,c,i;cin>>T;while(T--){cin>>n;int ll=1;memset(fs,0,sizeof(fs));memset(ans,0,sizeof(ans));memset(fss,0,sizeof(fss));for (i = 1; i <= n; ++i) node[i].l = node[i].r = node[i].w = -1;for(i=1;i<=n;i++){num[i].clear();scanf("%d",&w[i]);node[i].w=w[i];fss[ll++]=w[i];}cin>>m;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);node[a].l=b;node[a].r=c;}cin>>q;for(i=1;i<=q;i++){scanf("%d%d",&a,&b);fss[ll++]=b;xx.id=i;xx.x=b;num[a].push_back(xx);}len=1;sort(fss,fss+ll);for(i=1;i<ll;i++)if(fss[i]!=fss[i-1])fs[len++]=fss[i];dfs(1);for(i=1;i<=q;i++){if(ans[i][0]==-1)printf("0\n");else printf("%d %d\n",ans[i][0],ans[i][1]);}}return 0;}