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Problem Description In the battlefield , an effective way to defeat enemies is to break their communication system.
The information department told you that there are n enemy soldiers and their network which have n-1 communication routes can cover all of their soldiers. Information can exchange between any two soldiers by the communication routes. The number 1 soldier is the total commander and other soldiers who have only one neighbour is the frontline soldier.
Your boss zzn ordered you to cut off some routes to make any frontline soldiers in the network cannot reflect the information they collect from the battlefield to the total commander( number 1 soldier).
There is a kind of device who can choose some routes to cut off . But the cost (w) of any route you choose to cut off can’t be more than the device’s upper limit power. And the sum of the cost can’t be more than the device’s life m.
Now please minimize the upper limit power of your device to finish your task. Input The input consists of several test cases.
The first line of each test case contains 2 integers: n(n<=1000）m(m<=1000000).
Each of the following N-1 lines is of the form:
ai bi wi
It means there’s one route from ai to bi(undirected) and it takes wi cost to cut off the route with the device.
(1<=ai,bi<=n,1<=wi<=1000)
The input ends with n=m=0. Output Each case should output one integer, the minimal possible upper limit power of your device to finish your task.
If there is no way to finish the task, output -1. Sample Input 5 5 1 3 2 1 4 3 3 5 5 4 2 6 0 0 Sample Output 3 題解：在滿足總的花費不大於m的情況下，limit要盡量小，所以二分limit，然後檢查啦！（學過的東西不能融匯貫通，難怪我這麽菜，orz）

#include<bits/stdc++.h>
using
 
 namespace std;

const int maxn=1005;struct node{
    int to,next,va;
}e[2*maxn];

int n,m,tot;
int dp[maxn],head[maxn];

void Inite(){
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w){
    e[tot].to=v;
    e[tot].va=w;
    e[tot].next=head[u];
    head[u]=tot++;
}

void DFS(int
 
 pa,int u,int limit){
    dp[u]=0;
    int flag=0;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].to;
        if(pa==v) continue;
        flag=1;
        DFS(u,v,limit);
        
        if(e[i].va>limit) dp[u]+=dp[v];
        else dp[u]+=min(dp[v],e[i].va);
    }
    if(flag==0) dp[u]=100001
 
;
}

bool check(int limit){
    DFS(0,1,limit);
    if(dp[1]>m) return false;        
    else return true;
} 

void Solve(){
    int l=0,r=m+1;
    for(int i=1;i<=100;i++){
        int mid=(l+r)>>1;
        if(check(mid)) r=mid;
        else l=mid;
    }
    if(check(l)) printf("%d\n",l);
    else if(check(r)) printf("%d\n",r);
    else printf("-1\n");
}

int main()
{    
    while(~scanf("%d%d",&n,&m)){
        if(n==0&&m==0) break;
        
        Inite();
        for(int i=2;i<=n;i++){
            int u,v,va;
            scanf("%d%d%d",&u,&v,&va);
            addedge(u,v,va);
            addedge(v,u,va);
        }
        Solve();    
    }
    return 0;
}

Information Disturbing hdu-3586