题目地址：HDU 2242
先用双连通分量缩点，然后形成一棵树，然后在树上做树形DP，，求出每个点的子树和。然后找最小值即可。需要注意一下重边的问题，第一次返回父节点时可以忽略，因为这是反向边，然后之后再返回的时候就不是反向边了。不能忽略了。
代码如下：#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>using namespace std;#define LL long long#define pi acos(-1.0)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=10000+10;int head[MAXN], Ecnt, top, indx, scc;int low[MAXN], dfn[MAXN], belong[MAXN], stk[MAXN], dp[MAXN], a[MAXN];struct node{int u, v, next;}edge[MAXN<<2];void add(int u, int v){edge[Ecnt].u=u;edge[Ecnt].v=v;edge[Ecnt].next=head[u];head[u]=Ecnt++;}void tarjan(int u, int fa){low[u]=dfn[u]=++indx;stk[++top]=u;int flag=1;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(v==fa&&flag) {flag=0;continue ;}if(!dfn[v]){tarjan(v,u);low[u]=min(low[u],low[v]);}else low[u]=min(low[u],dfn[v]);}if(low[u]==dfn[u]){scc++;while(1){int v=stk[top–];belong[v]=scc;dp[scc]+=a[v];if(u==v) break;}}}void init(){memset(head,-1,sizeof(head));memset(dfn,0,sizeof(dfn));memset(dp,0,sizeof(dp));Ecnt=top=indx=scc=0;}int head1[MAXN], Ecnt1, vis[MAXN];struct node1{int u, v, next;}edge1[MAXN<<2];void add1(int u, int v){edge1[Ecnt1].v=v;edge1[Ecnt1].next=head1[u];head1[u]=Ecnt1++;}void dfs(int u){vis[u]=1;for(int i=head1[u];i!=-1;i=edge1[i].next){int v=edge1[i].v;if(vis[v]) continue ;dfs(v);dp[u]+=dp[v];}}void init1(){memset(head1,-1,sizeof(head1));memset(vis,0,sizeof(vis));Ecnt1=0;}int main(){int n, m, u, v, i, j, sum, min1;while(scanf("%d%d",&n,&m)!=EOF){sum=0;for(i=0;i<n;i++){scanf("%d",&a[i]);sum+=a[i];}init();while(m–){scanf("%d%d",&u,&v);add(u,v);add(v,u);}for(i=0;i<n;i++){if(!dfn[i]) tarjan(i,-1);}if(scc==1){puts("impossible");continue ;}init1();for(i=0;i<n;i++){for(j=head[i];j!=-1;j=edge[j].next){v=edge[j].v;if(belong[i]!=belong[v]){add1(belong[i],belong[v]);add1(belong[v],belong[i]);}}}dfs(1);min1=INF;for(i=1;i<=scc;i++){min1=min(min1,abs(sum-2*dp[i]));}printf("%d\n",min1);}return 0;}
每一个成功者都有一个开始。勇于开始，才能找到成功的路。
