HDU 3586.Information Disturbing 树形dp 叶子和根不联通的最小代价

Information Disturbing

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3205    Accepted Submission(s): 1137

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

 

Author

alpc86

 

Source

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

 

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题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=3586

题意：有一棵n个节点的树，现在需要删除一些边使得叶子节点和根（1）不是联通的，删除的总边权不超过m，并且最大边权最小。

思路：dp[u]表示u的子节点与u不相连的代价。<u,v>w，dp[u]+=min(dp[v],w)。当最大边权为x时，需要进行判断<u,v>w，如果w比x大的话，只能dp[u]+=dp[v]。所以二分最大边权。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<stack>
#include<set>
#include<bitset>
using namespace std;
#define PI acos(-1.0)
#define eps 1e-8
typedef long long ll;
typedef pair<int,int > P;
const int N=2e5+100,M=2e6+100;
const int inf=0x3f3f3f3f;
const ll INF=1e18+7,mod=1e9+7;
struct edge
{
    int from,to;
    ll w;
    int next;
};
edge es[M];
int cnt,head[N];
ll dp[N];
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,ll w)
{
    cnt++;
    es[cnt].from=u,es[cnt].to=v;
    es[cnt].w=w;
    es[cnt].next=head[u];
    head[u]=cnt;
}
void dfs(int u,int fa,ll mid)
{
    dp[u]=0LL;
    int child=0;
    for(int i=head[u]; i!=-1; i=es[i].next)
    {
        edge e=es[i];
        if(e.to==fa) continue;
        child++;
        dfs(e.to,u,mid);
        if(e.w>mid) dp[u]+=dp[e.to];
        else dp[u]+=min(dp[e.to],e.w);
    }
    if(!child) dp[u]=inf;
}
int main()
{
    int n;
    ll m;
    while(~scanf("%d%lld",&n,&m))
    {
        if(n==0&&m==0) break;
        init();
        ll l=inf,r=-inf;
        for(int i=1; i<n; i++)
        {
            int u,v;
            ll w;
            scanf("%d%d%lld",&u,&v,&w);
            l=min(l,w),r=max(r,w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        ll ans=0;
        while(l<=r)
        {
            ll mid=(l+r)>>1;
            dfs(1,0,mid);
            if(dp[1]<=m) r=mid-1,ans=mid;
            else l=mid+1;
        }
        if(ans) printf("%d
",ans);
        else printf("-1
");
    }
    return 0;
}