poj 3140 Contestants Division

Contestants Division

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7254 Accepted: 2056

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

Source

Shanghai 2006

题目大意给你一颗树。叫你找到一条边使得去掉这条边后形成的两棵子树的结点数相差最小。开始因为m的范围纠结半天按常识都知道m顶多为n-1的看了半天。结果还是硬着头皮写了。结果根本跟m没多大关系。。。目测数据有点大直接用__int64了。

#include <iostream>#include<algorithm>#include<string.h>#include<stdio.h>#define MAX(a,b) ((a)>(b)?(a):(b))#define positive(a) ((a)>0?(a):-(a))using namespace std;struct node1//点结构{    __int64 sum;//存子树的总结点数} points[100010];struct node2//边结构{    int to;//终点    node2 *next;} edge[1000010],*head[100010];__int64 psum[100010],temp,ans;int cnt,ptr,n,m;void adde(int f,int s)//加边{    edge[cnt].next=head[f];    edge[cnt].to=s;    head[f]=&edge[cnt++];}int ma(int a,int b){ return a>b?a:b; }void dfs(int fa,int son)//以s为根结点遍历树。计算每颗子树树的总结点数{    node2 *p=head[son];    while(p!=NULL)    {        if(p->to!=fa)        {            dfs(son,p->to);            points[son].sum+=points[p->to].sum;        }        p=p->next;    }    psum[ptr++]=points[son].sum;    //printf("%d %I64d\n",son,psum[ptr-1]);    //getchar();}void solve(int s){    int i;    ptr=0;    dfs(0,s);    temp=points[s].sum;    ans=100000000000000LL;    for(i=0;i<ptr-1;i++)    {        psum[i]=positive(temp-2*psum[i]);        if(psum[i]<ans)//把满足条件的结点存在数组中            ans=psum[i];    }    if(ptr==1)        ans=0;}int main(){    int i,a,b,cas=1;    while(scanf("%d%d",&n,&m),n||m)    {        memset(head,0,sizeof head);        cnt=0;        for(i=1;i<=n;i++)           scanf("%I64d",&points[i].sum);        for(i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            adde(a,b);            adde(b,a);        }        solve(1);        printf("Case %d: %I64d\n",cas++,ans);    }    return 0;}