POJ 3140 tree dp

Contestants Division

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:7725		Accepted:2196

Description

In the new ACM-ICPC Regional Contest, a special monitoring and submitting system will be set up, and students will be able to compete at their own universities. however there's one problem. due to the high cost of the new judging system, the organizing committee can only afford to set the system up such that there will be only one way to transfer information from one university to another without p Assing the same university twice. the contestants will be divided into two connected regions, and the difference between the total numbers of students from two regions shoshould be minimized. can you help the juries to find the minimum difference?

Input

There are multiple test cases in the input file. Each test case starts with two integersNAndM, (1 ≤N≤ 100000, 1 ≤M≤ 1000000), the number of universities and the number of direct communication line set up by the committee, respectively. Universities are numbered from 1N. The next line hasNIntegers,KTh integer is equal to the number of students in university numberedK. The number of students in any university does not exceed 100000000. Each of the followingMLines has two integersS,T, And describes a communication line connecting universitySAnd universityT. All communication lines of this new system are bidirectional.

N= 0,M= 0 indicates the end of input and shoshould not be processed by your program.

Output

For every test case, output one integer, the minimum absolute difference of students between two regions in the format as indicated in the sample output.

Sample Input

7 61 1 1 1 1 1 11 22 73 74 66 25 70 0

Sample Output

Case 1: 1

Source

Shanghai 2006

[Submit] [Go Back] [Status] [Discuss]

Divide a tree into two Subtrees so that the difference value is as small as possible.

It is easy to directly extract the weight of the subtree with each vertex as the root, then obtain the weight of the remaining part based on the total weight, and obtain the minimum value through enumeration update.

Code:

/* ***********************************************Author :xianxingwuguanCreated Time :2014-2-6 16:15:10File Name :1.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include 
 
  #include 
  
   #include #include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             using namespace std;#define INF 1000000000000000LL#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const ll maxn=100100;ll head[maxn],tol,weight[maxn],dp[maxn];struct node{ll next,to;node(){};node(ll _next,ll _to):next(_next),to(_to){}}edge[5*maxn];void add(ll u,ll v){edge[tol]=node(head[u],v);head[u]=tol++;}void dfs(ll u,ll fa){dp[u]=weight[u];for(ll i=head[u];i!=-1;i=edge[i].next){ll v=edge[i].to;if(v==fa)continue;dfs(v,u);dp[u]+=dp[v];}}ll ABS(ll x){if(x<=0)x=-x;return x;}int main(){ll i,j,k,m,n,p,T=0;while(cin>>n>>m){if(n==0&&m==0)break;memset(head,-1,sizeof(head));ll sum=0,ans=INF;for(i=1;i<=n;i++)cin>>weight[i],sum+=weight[i];while(m--){cin>>j>>k;add(j,k);add(k,j);}dfs(1,-1);for(i=1;i<=n;i++){ll pp=ABS(dp[i]-(sum-dp[i]));ans=min(ans,pp);}cout<<"Case "<<++T<<": "<