Contestants Division
Time Limit:2000MSMemory Limit:65536K
Total Submissions:8544Accepted:2439
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 integersNandM, (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 toN. The next line hasNintegers, theKthinteger is equal to the number of students in university numberedK. The number of students in any university does not exceed 100000000. Each of the followingMlines has two integerss,t, and describes a communication line connecting universitysand universityt. 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
类似 poj 2378
删边，求删去某条边后两个分支的最小差异值;注意数据范围。
#include<stdio.h>#include<math.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<queue>#include<vector>using namespace std;#define ll long long#define N 110000#define mem(a,t) memset(a,t,sizeof(a))const int inf=1000005;int cnt,n;struct node{int v,next;}e[N*20];int head[N];ll num[N];ll ans,sum;void add(int u,int v){e[cnt].v=v;e[cnt].next=head[u];head[u]=cnt++;}void dfs(int u,int len,int fa){int i,v;for(i=head[u];i!=-1;i=e[i].next){v=e[i].v;if(v!=fa){dfs(v,len+1,u);num[u]+=num[v];}}ll t=sum-2*num[u];if(t<0)t*=-1;if(t<ans)ans=t;}int main(){//freopen("in.txt","r",stdin);int i,u,v,m,cas=1;while(scanf("%d%d",&n,&m),n||m){for(i=1,sum=0;i<=n;i++){scanf("%lld",&num[i]);sum+=num[i];}cnt=0;mem(head,-1);for(i=0;i<m;i++){scanf("%d%d",&u,&v);add(u,v);add(v,u);}ans=sum;dfs(1,0,-1);printf("Case %d: %lld\n",cas++,ans);}return 0;}
，从起点，到尽头，也许快乐，或有时孤独，
