2013 多校联合 F Magic Ball Game （hdu 4605）

http://acm.hdu.edu.cn/showproblem.php?pid=4605

Magic Ball Game

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 309    Accepted Submission(s): 73

Problem Description

When the magic ball game turns up, Kimi immediately falls in it. The interesting game is made up of N balls, each with a weight of w[i]. These N balls form a rooted tree, with the 1st ball as the root. Any ball in the game has either 0 or 2 children ball. If a node has 2 children balls, we may define one as the left child and the other as the right child.
The rules are simple: when Kimi decides to drop a magic ball with a weight of X, the ball goes down through the tree from the root. When the magic ball arrives at a node in the tree, there's a possibility to be catched and stop rolling, or continue to roll down left or right. The game ends when the ball stops, and the final score of the game depends on the node at which it stops.
After a long-time playing, Kimi now find out the key of the game. When the magic ball arrives at node u weighting w[u], it follows the laws below:
1  If X=w[u] or node u has no children balls, the magic ball stops.
2  If X<w[u], there's a possibility of 1/2 for the magic ball to roll down either left or right.
3  If X>w[u], the magic ball will roll down to its left child in a possibility of 1/8, while the possibility of rolling down right is 7/8.
In order to choose the right magic ball and achieve the goal, Kimi wonders what's the possibility for a magic ball with a weight of X to go past node v. No matter how the magic ball rolls down, it counts if node v exists on the path that the magic ball goes along.
Manual calculating is fun, but programmers have their ways to reach the answer. Now given the tree in the game and all Kimi's queries, you're required to answer the possibility he wonders.

 

Input

The input contains several test cases. An integer T(T≤15) will exist in the first line of input, indicating the number of test cases.
Each test case begins with an integer N(1≤N≤10⁵), indicating the number of nodes in the tree. The following line contains N integers w[i], indicating the weight of each node in the tree. (1 ≤ i ≤ N, 1 ≤ w[i] ≤ 10⁹, N is odd)
The following line contains the number of relationships M. The next M lines, each with three integers u,a and b(1≤u,a,b≤N), denotes that node a and b are respectively the left child and right child of node u. You may assume the tree contains exactly N nodes and (N-1) edges.
The next line gives the number of queries Q(1≤Q≤10⁵). The following Q lines, each with two integers v and X(1≤v≤N,1≤X≤10⁹), describe all the queries.

 

Output

If the magic ball is impossible to arrive at node v, output a single 0. Otherwise, you may easily find that the answer will be in the format of 7^x/2^y . You're only required to output the x and y for each query, separated by a blank. Each answer should be put down in one line.

 

思路: 首先考虑一个询问（V,X），若从根节点到V节点的路径上（不包括V）已存在权值为X的节点，则小球不可能到达V节点，否则，不妨定义往左走的路径为“左路径”，往右走的路径称为“右路径”。设lmi，lma，rmi，rma分别表示左路径上小于X的节点数，左路径上大于X的节点数，右路径上小于X的节点数，右路径上大于X的节点数，则最终的答案即为： （1/2）^(lma+rma)*(7/8)^(rmi)*（1/8）^(lmi)。即输出 rmi 和 3*（lmi+rmi）+（lma+rma）即可。

  对于本题，我们可以离线处理每个节点上的询问，从根节点做一次DFS，在过程中每经过一个节点就处理该节点所对应的询问，我们可以用线段树，树状数组等数据结构记录从根节点到V的路径上所有的权值（不包括V），然后就可以得到lmi，lma，rmi，rma，剩下的就是更行答案了，另外X很大，需要离散化处理。

代码如下：

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <vector>#define maxn 100010#define mid ((t[p].l+t[p].r)>>1)#define ls (p<<1)#define rs (ls|1)using namespace std;struct Tree{    int w;    int left,right;}node[maxn];vector<int> vec[maxn];int c[maxn<<1][2];void init(int n){    for(int i=0;i<=n;i++)    {        vec[i].clear();        node[i].left=node[i].right=node[i].w=-1;    }}void add(int a,int b,int c){    node[a].left=b;    node[a].right=c;}int ans[maxn][2],vis[maxn],po[maxn<<1],len;int lowbit(int x){    return x&(-x);}void addnum(int x,int val,int tt){    while(x<=len)    {        c[x][tt]+=val;        x+=lowbit(x);    }}int getsum(int x,int tt){    int sum=0;    while(x>0)    {        sum+=c[x][tt];        x-=lowbit(x);    }    return sum;}int search(int len,int x){    int mi=1,ma=len,Mid;    while(mi<=ma)    {        Mid=(mi+ma)>>1;        if(po[Mid]==x)        return Mid;        if(po[Mid]<x)        mi=Mid+1;        else        ma=Mid-1;    }}void dfs(int now){    int i,w=search(len,node[now].w);    for(i=0;i<vec[now].size();i+=2)    {        int x=search(len,vec[now][i]),num=vec[now][i+1];        if(getsum(x,0)-getsum(x-1,0)>0||getsum(x,1)-getsum(x-1,1)>0)//如果已经存在x        {            ans[num][0]=-1;        }        else        {            int lma,lmi,rma,rmi;//   左边大于，左边小于，右边大于，右边小于            lmi=getsum(x-1,0);            lma=getsum(len,0)-getsum(x,0);            rmi=getsum(x-1,1);            rma=getsum(len,1)-getsum(x,1);            ans[num][0]=rmi;            ans[num][1]=3*(rmi+lmi)+rma+lma;        }    }    if(node[now].left!=-1)    {        addnum(w,1,0);        dfs(node[now].left);        addnum(w,-1,0);    }    if(node[now].right!=-1)    {        addnum(w,1,1);        dfs(node[now].right);        addnum(w,-1,1);    }}int main(){    //freopen("dd.txt","r",stdin);    int ncase;    scanf("%d",&ncase);    while(ncase--)    {        int n,i,m,q,x,v;        scanf("%d",&n);        init(n);        for(i=1;i<=n;i++)        {            scanf("%d",&node[i].w);            po[i]=node[i].w;        }        scanf("%d",&m);        int a,b,cc;        memset(vis,0,sizeof(vis));        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&cc);            add(a,b,cc);            vis[b]=vis[cc]=1;        }        scanf("%d",&q);        for(i=1;i<=q;i++)        {            scanf("%d%d",&v,&x);            vec[v].push_back(x);            vec[v].push_back(i);            po[i+n]=x;        }        sort(po+1,po+q+n+1);        len=unique(po+1,po+q+n+1)-(po+1);        int root;        for(i=1;i<=n;i++)        {            if(!vis[i])            {                root=i;                break;            }        }        for(i=0;i<=len;i++)        c[i][0]=c[i][1]=0;        dfs(root);        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;}