Contestants Division （树形dp）

最新推荐文章于 2020-03-15 22:51:32 发布

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最新推荐文章于 2020-03-15 22:51:32 发布

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分类专栏：动态规划树形dp 文章标签：动态规划树形dp 算法

本文链接：https://blog.youkuaiyun.com/a1046765624/article/details/78578119

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探讨在ACM-ICPC竞赛中如何通过优化算法寻找最佳的学生分组方案，确保两组间人数差异最小。文章提供了一种基于深度优先搜索（DFS）的方法来解决这一问题，并附带实现代码。

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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 6
1 1 1 1 1 1 1
1 2
2 7
3 7
4 6
6 2
5 7
0 0

Sample Output

Case 1: 1

题目大概：

在一个树中，每个结点都有价值，找到两个连通区域，价值差距最小，是多少。

思路：

就是以一个节点为分界线，他的儿子往下走是一个区域，它的父亲算一个区域，计算两个区域的差值。

这样的话，就dfs一边，先更新儿子，再利用儿子更新父亲，计算差值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define ll long long
const int ma=100005;
ll n,m;
ll dp[ma];
ll nao[ma];
ll head[ma];
ll ans,sum;
ll minshu;
struct shu
{
    int v;
    int next;
}tr[2000010];

void add(ll q,ll w)
{
    tr[ans].v=w;
    tr[ans].next=head[q];
    head[q]=ans++;

}

ll f(ll a)
{
    if(a<0)return -a;
    else return a;
}
void dfs(int x,int pa)
{
    dp[x]=nao[x];
    for(int i=head[x];i!=-1;i=tr[i].next)
    {
        int son=tr[i].v;
        if(pa!=son)
        {
            dfs(son,x);
            dp[x]+=dp[son];
        }
    }
    minshu=min(f(2*dp[x]-sum),minshu);
}

int main()
{

   int t=1;
   while(~scanf("%I64d%I64d",&n,&m))
   {
       if(n==0&&m==0)break;
       memset(dp,0,sizeof(dp));
       memset(head,-1,sizeof(head));
       memset(nao,0,sizeof(nao));
       ans=0;
       sum=0;
       for(int i=1;i<=n;i++)
       {
           scanf("%I64d",&nao[i]);
           sum+=nao[i];
       }
       for(int i=1;i<=m;i++)
       {
           int q,w;
            scanf("%d%d",&q,&w);
            add(q,w);
            add(w,q);

       }
       minshu=sum;
       dfs(1,-1);

       printf("Case %d: %I64d\n",t,minshu);
       t++;
   }
    return 0;
}