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poj3140 Contestants Division（樹形dp）

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poj3140 Contestants Division（樹形dp）

Contestants Division Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9004 Accepted: 2583

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 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

Source

Shanghai 2006
題意：給定一顆n節點的樹，每個結點有k個學生；求刪除一條邊之後分成的兩棵子樹的學生數差最小，輸出差值。分析：求出每個結點以自己為根的樹的節點數，然後用總學生數去減就行了。 Ps：這題真的讓我心碎，WA無數遍，但一直找不到錯，最後對照著某大牛博客修改，依然WA，最後我煩的很，把所有不一樣的全部改了，終於找出了錯誤，（見代碼）然而我並不知道為什麼。如果你知道的話，請告訴我，please！

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
#define INF 21474836470000000
//const int INF = 0x3f3f3f3f;   //提交到心碎，結果是這裡的問題
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 200010
#define JJ(a)  (a<0?-a:a)
#define min(a,b) (a)<(b)?(a):(b)

struct node
{
    int v;
    node *next;
}tree[MAXN], *head[MAXN];
int n,m,pre,num[MAXN],vis[MAXN];
ll ans,dp[MAXN],sum;

void init()
{
    sum = 0;
    pre = 1;
    ans = INF;
    CL(vis, 0);
    CL(head, NULL);
}

void add(int x, int y)
{
    tree[pre].v = y;
    tree[pre].next = head[x]; head[x] = &tree[pre++];
    tree[pre].v = x;
    tree[pre].next = head[y]; head[y] = &tree[pre++];
}

void dfs(int son, int father)//求出每個結點為根包含的學生數
{
    dp[son] = num[son];
    vis[son] = 1;
    node *p = head[son];
    while(p != NULL)
    {
        if(vis[p->v] == 0)
        {
            dfs(p->v, son);
            dp[son] += dp[p->v];
        }
        p = p->next;
    }
}

void dp_tree()//直接暴力求就行了
{
    for(int i=1; i<=n; i++)
    {
        ll tp = dp[i];
        ans = min(ans, JJ((2*tp-sum)));
    }
}

int main()
{
    int a,b,cas=1;
    while(scanf("%d%d",&n,&m),n+m)
    {
        init();
        for(int i=1; i<=n; i++)
            scanf("%d",&num[i]), sum += num[i];
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            add(a, b);
        }
        dfs(1, 0);
        CL(vis, 0);
        dp_tree();
        printf("Case %d: %I64d\n",cas++,ans);
    }
    return 0;
}