hdu 2242 (双联通+树形DP)

Tarjan双联通缩点后,建树,任选一点为根节点求出所有点的字节点的个数+1:m

然后求出n-m与m的差值,求出最小的






#include<stdio.h>
#include<stack>
#include<string.h>
#define inf 0x3fffffff
#define N 10001
using namespace std;
int belong[N],dfs[N],low[N],idx,ans,head[N],num,n,m,people[N],pe[N],nume,sum,vis[N];
struct edge
{
	int st,ed,next;
}E[N*4],e[N*4];
void addedge(int x,int y)
{
	E[num].st=x;
	E[num].ed=y;
	E[num].next=head[x];
	head[x]=num++;
}
void Addedge(int x,int y)
{
	e[nume].st=x;
	e[nume].ed=y;
	e[nume].next=head[x];
	head[x]=nume++;
}
stack<int>Q;
void Tarjan(int u,int father)
{
    int i,j,v,flag=0;
	low[u]=dfs[u]=idx++;
	Q.push(u);
	for(i=head[u];i!=-1;i=E[i].next)
	{
		v=E[i].ed;
		if(dfs[v]==-1)
		{
			Tarjan(v,u);
			low[u]=low[u]>low[v]?low[v]:low[u];
		}
		else if(v==father)
		{
			if(flag)
				low[u]=low[u]>dfs[v]?dfs[v]:low[u];
			flag++;
		}
		else low[u]=low[u]>dfs[v]?dfs[v]:low[u];
	}
	if(dfs[u]==low[u])
	{
		do
		{
			j=Q.top();
			Q.pop();
			belong[j]=ans;
			people[ans]+=pe[j];
		}while(j!=u);
		ans++;
	}
}
void buildmap()
{
	memset(head,-1,sizeof(head));
	nume=0;
	for(int i=0;i<num;i+=2)
	{
       int x=belong[E[i].st];
	   int y=belong[E[i].ed];
	   if(x==y)continue;
	   Addedge(x,y);
	   Addedge(y,x);
	}
}
int mn;
int discost(int u)
{
	int cont=people[u];
	vis[u]=1;
	for(int i=head[u];i!=-1;i=e[i].next)
	{
		int v=e[i].ed;
		if(vis[v]==1)continue;
		int temp=discost(v);
		cont+=temp;
		temp=sum-2*temp;
		if(temp<0)
			temp=-temp;
		if(mn>temp)
			mn=temp;
	}
	return cont;
}
int main()
{
	int i,x,y;
	while(scanf("%d%d",&n,&m)!=-1)
	{
		sum=0;
		memset(head,-1,sizeof(head));
		num=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&pe[i]);
			sum+=pe[i];
		}
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&x,&y);
			addedge(x,y);
			addedge(y,x);
		}
		idx=ans=0;
		memset(dfs,-1,sizeof(dfs));
		memset(people,0,sizeof(people));
		Tarjan(0,-1);
		if(ans==1)
        {
            printf("impossible\n");
            continue;
        }
		buildmap();
		memset(vis,0,sizeof(vis));
		mn=inf;
		discost(0);
		printf("%d\n",mn);
	}
	return 0;
}


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