POJ1741 Tree

POJ1741 Tree
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 6232   Accepted: 1770

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 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

Sample Output

8
----------------------------------------------------------------
题目大意:见漆子超论文。
解题思路:见漆子超论文。表示一A,爽歪歪,原来是这样子的,还是觉得挺暴力的,只是避免了最坏情况而已。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;

const int N=10005,M=20005,inf=0x3f3f3f3f;
int n,lim,eid,minn;
int head[N],ed[M],val[M],nxt[M];
int vis[N],fa[N],siz[N],dep[N],le,ri;

void addedge(int s,int e,int v){
    ed[eid]=e;val[eid]=v;nxt[eid]=head[s];head[s]=eid++;
}

int dfssize(int s,int f){
    fa[s]=f;siz[s]=1;
    for(int i=head[s];~i;i=nxt[i])
        if(ed[i]!=f&&!vis[ed[i]])siz[s]+=dfssize(ed[i],s);
    return siz[s];
}

void dfsroot(int s,int sum,int&root){
    int maxx=sum-siz[s];
    for(int i=head[s];~i;i=nxt[i]){
        int e=ed[i];if(e==fa[s]||vis[e])continue;
        dfsroot(e,sum,root);maxx=max(maxx,siz[e]);
    }
    if(maxx<minn){minn=maxx;root=s;}
}

void dfsdepth(int s,int d,int f){
    dep[ri++]=d;
    for(int i=head[s];~i;i=nxt[i])
        if(ed[i]!=f&&!vis[ed[i]])dfsdepth(ed[i],d+val[i],s);
}

int getdep(int a,int b){
    sort(dep+a,dep+b);int ret=0,e=b-1;
    for(int i=a;i<b;i++){
        if(dep[i]>lim)break;
        while(e>=a&&dep[e]+dep[i]>lim)e--;
        ret+=e-a+1;if(e>i)ret--;
    }
    return ret>>1;
}

int solve(int s){
    int sum=dfssize(s,-1),ret=0;minn=inf;
    int root;dfsroot(s,sum,root);
    vis[root]=1;
    for(int i=head[root];~i;i=nxt[i])
        if(ed[i]!=root&&!vis[ed[i]])
            ret+=solve(ed[i]);
    le=ri=0;
    for(int i=head[root];~i;i=nxt[i])
        if(ed[i]!=root&&!vis[ed[i]]){
            dfsdepth(ed[i],val[i],-1);ret-=getdep(le,ri);le=ri;
        }
    ret+=getdep(0,ri);
    for(int i=0;i<ri;i++)
        if(dep[i]<=lim)ret++;
        else break;
//    printf("%d %d %d %d\n",root,ret,le,ri);
    vis[root]=0;
    return ret;
}

int main(){
//    freopen("/home/axorb/in","r",stdin);
    while(scanf("%d%d",&n,&lim),n+lim){
        eid=0;clr(head,-1);
        for(int i=1;i<n;i++){
            int a,b,c;scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,c);addedge(b,a,c);
        }
        clr(vis,0);
        printf("%d\n",solve(1));
    }
}

  

也许有挫折,但这些,怎能挡住湘北前进的步伐
原文地址:https://www.cnblogs.com/Fatedayt/p/2580555.html