POJ-3140-Contestants Division

Contestants Division

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10926 Accepted: 3058

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 passing 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 should 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 integersN and M, (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 1 toN. The next line has N integers, the K^th 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 followingM lines has two integers s, t, and describes a communication line connecting universitys and university t. All communication lines of this new system are bidirectional.

N = 0, M = 0 indicates the end of input and should 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

题意：求删除一个点后的形成的两个子树的节点的差值。

因为是二叉树嘛更好做了，求出一个子树的节点数，然后差值就是|总结点数-2*任一子树节点数目|。

代码：

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>#include<vector>using namespace std;int n,m;vector<int>tree[100005];int v[100005];long long minn,s;void csh(){for(int i=0;i<=n;i++)    tree[i].clear();minn=9223372036854775807;s=0;}void build(int fa,int son){fa--;son--;tree[fa].push_back(son);tree[son].push_back(fa);}long long dfs(int pp,int to){long long ans=0;long long sum=0;for(int i=0;i<tree[pp].size();i++)    {    if(tree[pp][i]==to)        continue;    long long ss=tree[pp][i];    ans=dfs(ss,pp);    long long c=s-2*ans;    if(c<0)        c*=-1;    if(minn>c)        minn=c;    sum+=ans;    }return sum+v[pp];}int main(){int j=1,a,b;while(cin>>n>>m&&n+m)    {    csh();    for(int i=0;i<n;i++)        {        scanf("%d",&v[i]);        s+=v[i];        }    for(int i=0;i<m;i++)        {        scanf("%d%d",&a,&b);        build(a,b);        }    dfs(0,-1);    cout<<"Case "<<j++<<": ";    cout<<minn<<endl;    }return 0;}