Contestants Division POJ - 3140

探讨在高成本约束下,如何优化ACM-ICPC区域赛的监测与提交系统,通过将参赛学生分为两个连接区域并最小化两区域学生总数差异,确保竞赛公平与高效。本文介绍了一种基于树形结构的算法解决方案,通过深度优先搜索确定最佳边删除位置,以实现两部分学生数量的最小区别。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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 Kth 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 st, 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

题意:给你一棵树,树的每个节点都有一个权值,问你删掉一条边,使得分成的两棵树权值之和相差最小。

 

思路:dfs搜每个点的子树的权值之和,设为tmp,若从这个点的上一条边断开,则分成两个数的权值是tmp和sum-tmp,

那么他们的差为abs(sum-2*tmp);我们用ans维护一个最小值即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=100009;
const int maxm=1000009;
struct node{
	int id;
	int next;
}side[maxm*2];
int head[maxn],cnt,n,m;
ll val[maxn],sum,ans;
void init()
{
	memset(head,-1,sizeof(head));
	cnt=sum=0;
}
ll Abs(ll x)
{
	if(x<=0)
		return -x;
	else
		return x;
}
void add(int x,int y)
{
	side[cnt].id=y;
	side[cnt].next=head[x];
	head[x]=cnt++;
}
ll dfs(int x,int fa)
{
	ll tmp=val[x];
	for(int i=head[x];i!=-1;i=side[i].next)
	{
		int y=side[i].id;
		if(y==fa) continue;
		tmp+=dfs(y,x);
	}
	ans=min(ans,Abs(sum-2*tmp));
	return tmp;
}
int main()
{
	int x,y,cas=1;
	while(scanf("%d%d",&n,&m)&&(m+n))
	{
		init();
		for(int i=1;i<=n;i++)
		{
			scanf("%lld",&val[i]);
			sum+=val[i];	
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			add(x,y);
			add(y,x);
		}
		ans=sum;
		dfs(1,-1);
		printf("Case %d: %lld\n",cas++,ans);
	}
	return 0;
}

 

### 关于PAT编程能力测试排名汇总 对于PAT编程能力测试中的排名汇总,在C语言版本的具体实现中,可以考虑如下方式来处理成绩统计与排名计算。通常情况下,这类问题涉及读取输入数据、解析这些数据并按照特定规则进行排序最后输出结果。 考虑到PAT平台可能存在的某些特性或限制,比如编辑器不支持`gets()`函数[^1],因此建议采用更安全可靠的输入方法如`scanf()`来进行字符串或其他类型的输入操作。 下面是一个简单的C语言代码片段用于模拟PAT排名汇总的功能: ```c #include <stdio.h> #include <stdlib.h> // 定义结构体存储参赛者信息 typedef struct { int id; float score; } Contestant; // 比较函数供qsort使用 int compare(const void *a, const void *b) { return ((Contestant *)b)->score - ((Contestant *)a)->score; } int main() { int N; // 参赛人数 scanf("%d", &N); Contestant contestants[N]; for (int i = 0; i < N; ++i) { contestants[i].id = i + 1; scanf("%f", &(contestants[i].score)); } qsort(contestants, N, sizeof(Contestant), compare); printf("Ranking:\n"); for (int i = 0; i < N; ++i) { printf("ID:%d Score:%.2f\n", contestants[i].id, contestants[i].score); } return 0; } ``` 此段代码实现了基本的成绩录入和基于分数高低的排序功能,并打印出最终排名情况。需要注意的是实际比赛中可能会有更加复杂的要求,例如相同分数如何处理等问题,则需进一步调整逻辑满足具体需求[^2]。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值