POJ P1741 Tree

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题目

Tree

Time Limit: 1000MS Memory Limit: 30000K

Description

Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input

The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
Output

For each test case output the answer on a single line.
Sample Input

5 41 2 31 3 11 4 23 5 10 0

Sample Output

8

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题目大意

给出一个有n个顶点的树，每条边都有一个长度(正整数小于1001)。

定义dist(u,v)=节点u和v之间的最小距离。

给出一个整数k，对每对(u,v)顶点都是有效的，如果且仅当dist(u,v)不超过k。

编写一个程序，该程序将计算对给定树有效的对数对数。

输入

输入包含几个测试用例。每个测试用例的第一行包含两个整数n,k。(n < = 10000)下面的n - 1行每个包含三个整数u,v,l，这意味着节点u和长度l的v之间有一条边。

最后一个测试用例后面是两个0。

输出

对于每个测试用例输出一行，为dist(u,v)不超过k的u,v的对数。

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题解

点分治

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代码

#include<cstdio>#include<iostream>#include<cstdlib>#define maxn 10005#define INF 0x7fffffffusing namespace std;int n,m,tot,rt,size,ans;int lnk[maxn],d[maxn],vis[maxn],f[maxn],son[maxn],dp[maxn];struct edge{    int v,y,nxt;} e[maxn*2];int readln(){    int x=0;    char ch=getchar();    while (ch<'0'||ch>'9') ch=getchar();    while ('0'<=ch&&ch<='9') x=x*10+ch-48,ch=getchar();    return x;}void insert(int x,int y,int v){    tot++;    e[tot].nxt=lnk[x];    lnk[x]=tot;    e[tot].v=v;    e[tot].y=y;}void getrt(int x,int fa){    f[x]=0;son[x]=1;    for (int i=lnk[x];i;i=e[i].nxt)    {        if (e[i].y==fa||vis[e[i].y]) continue;        getrt(e[i].y,x);        son[x]+=son[e[i].y];        f[x]=max(f[x],son[e[i].y]);    }    f[x]=max(f[x],size-son[x]);    if (f[x]<f[rt]) rt=x;}void getdp(int x,int fa){    dp[0]++;    dp[dp[0]]=d[x];    for (int i=lnk[x];i;i=e[i].nxt)    {        if (e[i].y==fa||vis[e[i].y]) continue;        d[e[i].y]=d[x]+e[i].v;        getdp(e[i].y,x);    }}void qsort(int l,int r){    int i=l,j=r,mid=dp[rand()%(r-l+1)+l],t;    do {        while (dp[i]<mid) i++;        while (dp[j]>mid) j--;        if (i<=j) {            t=dp[i];dp[i]=dp[j];dp[j]=t;            i++;j--;        }    } while (i<=j);    if (i<r) qsort(i,r);    if (l<j) qsort(l,j);}int cal(int x,int v){    d[x]=v;dp[0]=0;    getdp(x,0);    qsort(1,dp[0]);    int l=1,r=dp[0],ans=0;    while (l<=r)     if (dp[l]+dp[r]<=m) {        ans+=r-l;        l++;     }     else r--;     return ans;}void solve(int x){    ans+=cal(x,0);    vis[x]=1;    for (int i=lnk[x];i;i=e[i].nxt)    {        if (vis[e[i].y]) continue;        ans-=cal(e[i].y,e[i].v);        rt=0;        size=son[e[i].y];        getrt(e[i].y,0);        solve(rt);    }}int main(){    while (true)    {        ans=0;rt=0;tot=0;size=0;f[0]=INF;        for (int i=1;i<=maxn;i++)        {            vis[i]=0;            lnk[i]=0;        }        n=readln();m=readln();        if (n==0&&m==0) break;        for (int i=1;i<n;i++)        {            int x=readln(),y=readln(),v=readln();            insert(x,y,v);insert(y,x,v);        }        getrt(1,0);        solve(rt);        printf("%d\n",ans);    }}