树链剖分-点的分治（dis[i]+dis[j]==k的点对数量）

poj2114

Boatherds

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1195		Accepted: 387

Description

Boatherds Inc. is a sailing company operating in the country of Trabantustan and offering boat trips on Trabantian rivers. All the rivers originate somewhere in the mountains and on their way down to the lowlands they gradually join and finally the single resulting river flows to the sea. Moreover, the Trabantian villages are exactly at the rivers' springs, junctions and at the mouth of the largest river. Please note that more than 2 rivers can join at a junction. However, the rivers always form a tree (with villages as vertices). 

The pricing policy of the Boatherds is very simple: each segment of each river between two villages is assigned a price (the price is same in both directions), so if a tourist requests a journey between any two villages, the ticket office clerks just add the prices of the segments along the only path between the villages. 

One day, a very strange tourist appeared. She told the clerks that she returns to her country on the next day and she wants to spend all the remaining money on a boat trip, so they should find a route with exactly this cost. Being just poor (ahem) businessmen, they have asked the Abacus Calculator Makers for help. 

You are given a description of the river network with costs of river segments and a sequence of integers x1,..., xk. For each xi, you should determine if there is a pair of cities (a, b) in the river network such that the cost of the trip between a and b is exactly xi. 

Input

The input consists of several instances. Each instance is described by (in the following order): 

• A single line containing a single integer: the number of villages N (1 <= N <= 10 000). 
  
• N lines describing the villages. The i-th of these lines (1 <= i <= N) describes the village with number i. It contains space separated integers d1, c1, d2, c2, , d_ki, c_ki, 0. The dj's are numbers of villages from which the rivers flow directly to the village i (with no other villages in between), each cj is the price of the journey between villages i and dj. Moreover, 2 <= dj <= N and 0 <= cj <= 1 000. Village 1 always corresponds to the mouth of the largest river, therefore no di can ever be equal to 1. 
  
• M <= 100 lines describing the queries. The i-th of these lines corresponds to the i-th query and contains a single integer xi (1 <= xi <= 10 000 000). 
  
• The instance is finished by a single line containing the number 0.

The whole input is ended by a single line containing the number 0. 

Output

For each instance you should produce a sequence of M lines (where M is the number of queries in the particular instance). The i-th of these lines contains the word "AYE" if there exists a pair of cities in the river network which is connected by a path of cost xi, or the word "NAY" otherwise. 

Output for each instance must be followed by a single line containing just the dot character. 

Sample Input

6
2 5 3 7 4 1 0
0
5 2 6 3 0
0
0
0
1
8
13
14
0
0

Sample Output

AYE
AYE
NAY
AYE
.

题目大意：

求一棵树上是否存在路径长度为K的点对。

POJ 1714求得是路径权值<=K的路径条数，这题只需要更改一下统计路径条数的函数即可，如果最终的路径条数大于零，则说明存在这样的路径。

刚开始我以为只要在分治过程中出现过长度为K的就算是找到了，其实不然，因为可能是相同子树里面的两个结点，这个结果显然是错误的。

修改内容：例如一个序列0,1,2,2,3,3,3,4,4,4,6,8,9

设k=6对于该子树，先找到0,1,2,2,3,3,3,4,4,4,6，sum+=1*1，

然后：搜到2,2,3,3,3,4,4,4，sum+=2*3（2个2与3个4）

最后搜到3,3,3，sum+=3*2/2;

return sum=9;

程序：

#include"string.h"
#include"stdio.h"
#include"stdlib.h"
#include"queue"
#include"stack"
#include"iostream"
#include"algorithm"
#include"vector"
#define inf 1000000000
#define M 51111
using namespace std;
struct node
{
    int u,v,w,next;
}edge[M*3];
int t,head[M],use[M],dis[M],son[M],limit[M],k,cnt,MN,ID,ans;
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].w=w;
    edge[t].next=head[u];
    head[u]=t++;
}
void dfs_size(int u,int f)
{
    son[u]=1;
    limit[u]=0;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(f!=v&&!use[v])
        {
            dfs_size(v,u);
            son[u]+=son[v];
            limit[u]=max(limit[u],son[v]);
        }
    }
}
void dfs_root(int root,int u,int f)
{
    if(son[root]-son[u]>limit[u])
        limit[u]=son[root]-son[u];
    if(MN>limit[u])
    {
        ID=u;
        MN=limit[u];
    }
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(f!=v&&!use[v])
            dfs_root(root,v,u);
    }
}
void dfs_dis(int u,int f,int id)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(f!=v&&!use[v])
        {
            dfs_dis(v,u,id+edge[i].w);
        }
    }
    dis[cnt++]=id;
}
int cal(int u,int f,int id)
{
    cnt=0;
    int sum=0,i,j;
    dfs_dis(u,f,id);
    sort(dis,dis+cnt);
    i=0;
    j=cnt-1;
    while(i<j)//需要特殊注意的地方
    {
        if(dis[i]+dis[j]<k)
            i++;
        else if(dis[i]+dis[j]>k)
            j--;
        else
        {
            if(dis[i]==dis[j])
            {
                sum+=(j-i+1)*(j-i)/2;
                break;
            }
            int st=i,ed=j;
            while(dis[st]==dis[i])st++;
            while(dis[ed]==dis[j])ed--;
            sum+=(st-i)*(j-ed);
            i=st,j=ed;
        }
    }
    return sum;
}
void dfs_ans(int root,int u,int f)
{
    dfs_size(root,f);
    MN=inf;
    dfs_root(root,root,f);
    ans+=cal(ID,ID,0);
    use[ID]=1;
    for(int i=head[ID];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!use[v])
        {
            ans-=cal(v,v,edge[i].w);
            dfs_ans(v,v,v);
        }
    }
}
void slove()
{
    ans=0;
    memset(use,0,sizeof(use));
    dfs_ans(1,1,1);
    //printf("%d\n",ans);
    if(ans>0)
        printf("AYE\n");
    else
        printf("NAY\n");
}
int main()
{
    int n,i,j,w;
    while(scanf("%d",&n),n)
    {
        init();
        for(i=1;i<=n;i++)
        {
            while(scanf("%d",&j),j)
            {
                scanf("%d",&w);
                add(i,j,w);
                add(j,i,w);
            }
        }
        while(scanf("%d",&k),k)
        {
            slove();
        }
        printf(".\n");
    }
    return 0;
}