Contestants Division - POJ 3140 树形dp

Contestants Division

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7987 Accepted: 2280

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 integers N 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 to N. The next line has N integers, the K^th integer is equal to the number of students in university numbered K. The number of students in any university does not exceed 100000000. Each of the following M lines has two integers s, t, and describes a communication line connecting university s 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

题意：给你一棵树，让你将这棵树分成两组，问权值的差最小是多少。

思路：dfs，比较简单就不解释了。另外我有一点困惑的是边的数量比较多，应该存在形成环的情况，可是我去搜题解的时候发现都没有说这一点，好像都是按照一棵树来做的，然后我就试了下，过了……欢迎大家批评指教。

莫名其妙的AC代码如下：

#include<cstdio>#include<cstring>#include<vector>using namespace std;vector<int> vc[100010];long long value[100010],sum,ans;int n,m,vis[100010];void dfs(int u){ int i,j,k,v;  vis[u]=1;  for(i=0;i<vc[u].size();i++)  { v=vc[u][i];    if(vis[v]==0)    { dfs(v);      value[u]+=value[v];      if(value[v]*2-sum<0)       ans=min(ans,sum-value[v]*2);      else       ans=min(ans,value[v]*2-sum);    }  }}int main(){ int i,j,k,u,v,t=0;  while(~scanf("%d%d",&n,&m) && n+m)  { sum=0;    ans=1e18;    for(i=1;i<=n;i++)    { vc[i].clear();      vis[i]=0;    }    for(i=1;i<=n;i++)    { scanf("%I64d",&value[i]);      sum+=value[i];    }    for(i=1;i<=m;i++)    { scanf("%d%d",&u,&v);      vc[u].push_back(v);      vc[v].push_back(u);    }    dfs(1);    printf("Case %d: %I64d\n",++t,ans);  }}