題目描述

Information Disturbing
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4003 Accepted Submission(s): 1391

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

大意

題目大意：給定n個(gè)敵方據(jù)點(diǎn)，1為司令部，其他點(diǎn)各有一條邊相連構(gòu)成一棵樹，每條邊都有一個(gè)權(quán)值cost表示破壞這條邊的費(fèi)用，葉子節(jié)點(diǎn)為前線。現(xiàn)要切斷前線和司令部的聯(lián)系，每次切斷邊的費(fèi)用不能超過上限limit，問切斷所有前線與司令部聯(lián)系所花費(fèi)的總費(fèi)用少于m時(shí)的最小limit。1<=n<=1000,1<=m<=1000000

思路

• 二分+樹形dp
  
  
• 二分可能的limit,再判斷該limit下的花費(fèi)是否符合m;
• dp: 對(duì)于一個(gè)節(jié)點(diǎn)u及其子節(jié)點(diǎn)v,要切斷u子樹下所有葉子節(jié)點(diǎn)與u的聯(lián)系,有兩種方式.
• 設(shè)dp[u]表示切斷u子樹下所有葉子節(jié)點(diǎn)與u的聯(lián)系的最小花費(fèi)

不切斷u->v,$dp[u]+=dp[v]$;

切斷u->v,$dp[u]+=dis(u,v)$;

代碼

#include<cstdio> #include<string> #include<vector> #include<cstring> #include<iostream> #define re register int using namespace std; const int maxn=1e3+50; inline int read(){int x=0,w=1;char ch=getchar();while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();if(ch=='-') w=-1,ch=getchar();while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-48,ch=getchar();return x*w; } int head[maxn<<1]; long long dp[maxn]; int tot=1,INF=1e6; struct data{int to,nxt;long long w; }edge[maxn<<1]; void DFS(int u,int fa,int lim){int flag=0;dp[u]=0;for(int i=head[u];i;i=edge[i].nxt){int v=edge[i].to;if(v!=fa){flag=1;DFS(v,u,lim);if(edge[i].w<=lim)dp[u]+=min(dp[v],edge[i].w);elsedp[u]+=dp[v];}} if(!flag) dp[u]=INF; } inline void add(int u,int v,int w){edge[tot].to=v;edge[tot].w=w;edge[tot].nxt=head[u];head[u]=tot++; } int main(){int MAXX=-1;int n,m;while(~scanf("%d%d",&n,&m)) { // cout<<n<<m<<endl;if(n==0&&m==0) break;tot=1;memset(head,0,sizeof(head));for(re i=1;i<n;++i) {int a,b,w;a=read(),b=read(),w=read();add(a,b,w);add(b,a,w);MAXX=max(MAXX,w);}int l=0,r=MAXX,ans=-1,mid;while(l<=r) {//cout<<endl;mid=(l+r)>>1;DFS(1,-1,mid);//cout<<l<<" "<<r<<endl;//cout<<dp[1]<<endl;if(dp[1]<=m){ans=mid;r=mid-1;}else l=mid+1;}printf("%d\n",ans);}return 0; }/* 5 5 1 3 2 1 4 3 3 5 5 4 2 6 0 0 */

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轉(zhuǎn)載于:https://www.cnblogs.com/bbqub/p/9073077.html

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