poj3140 樹狀dp

Contestants Division

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7396		Accepted: 2090

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 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 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

題目的意思就是，對於，一個樹，樹的每一個結點都有一個值，從一個結點把樹分開，要求兩部分所得的兩棵樹的差值是最小的，很明顯，就是樹狀dp！很簡單，但要注意，用__64才行！

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define MAXN 100005__int64 minxx,down[MAXN],prime[MAXN],sum;int n,head[MAXN],list[10*MAXN],next[10*MAXN],e,visit[MAXN];void addedge(int a,int b){    list[e]=b;next[e]=head[a];head[a]=e++;    list[e]=a;next[e]=head[b];head[b]=e++;}__int64 fmin(__int64 a,__int64 b){    if(a<b)    return a;    return b;}__int64 ffabs(__int64 a){    if(a>=0)    return a;    return -a;}void dfs(int x){   int i,temp;   visit[x]=1;   down[x]=prime[x];   for(i=head[x];i!=-1;i=next[i])   {       temp=list[i];       if(!visit[temp])       {           dfs(temp);           down[x]+=down[temp];       }   }    minxx=fmin(minxx,ffabs(sum-2*down[x]));}int main(){    int m,i,a,b,tcase=1;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0&&m==0)        break;        sum=0;        for(i=1;i<=n;i++)        {            scanf("%I64d",&prime[i]);            sum+=prime[i];        }        e=0;        memset(list,-1,sizeof(list));        memset(head,-1,sizeof(head));        memset(next,-1,sizeof(next));        memset(visit,0,sizeof(visit));        memset(down,0,sizeof(down));        for(i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            addedge(a,b);        }        minxx=sum;        dfs(1);        printf("Case %d: %I64d\n",tcase++,minxx);    }    return 0;}