BNUOJ 7697 Information Disturbing

Information Disturbing

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on HDU. Original ID: 3586
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

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

 

 

解题：要求什么寿命之和不能达到最大的寿命值，还要使得那个什么最小。。。最小。。。二分+树形dp。dp[u]表示u节点子树的最小代价。

 

dp[u] += min(dp[g[u][i].to],g[u][i].w);  表示要么不剪g[u][i]这条边，由其子树中的剪，要么剪g[u][i]这条边，子树不减，故取min就是。

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 1000010
using namespace std;
const int maxn = 1001;
struct arc{
    int to,w;
    arc(int x = 0,int y = 0):to(x),w(y){}
};
vector<arc>g[maxn];
int n,m,dp[maxn];
void dfs(int u,int fa,int limit){
    for(int i = 0; i < g[u].size(); i++){
        if(g[u][i].to == fa) continue;
        dfs(g[u][i].to,u,limit);
        if(g[u][i].w <= limit)
            dp[u] += min(dp[g[u][i].to],g[u][i].w);
        else dp[u] += dp[g[u][i].to];
    }
    if(g[u].size() == 1 && g[u][0].to == fa) dp[u] = INF;
}
int main() {
    int i,j,u,v,w,lt,rt,mid,ans;
    while(scanf("%d %d",&n,&m),n||m){
        for(i = 0; i <= n; i++)
            g[i].clear();
        for(lt = rt = 1,i = 1; i < n; i++){
            scanf("%d %d %d",&u,&v,&w);
            g[u].push_back(arc(v,w));
            g[v].push_back(arc(u,w));
            if(w > rt) rt = w;
        }
        ans = -1;
        while(lt <= rt){
            mid = (lt+rt)>>1;
            memset(dp,0,sizeof(dp));
            dfs(1,-1,mid);
            if(dp[1] <= m){
                ans = mid;
                rt = mid-1;
            }else lt = mid+1;
        }
        printf("%d
",ans);
    }
    return 0;
}