poj 3140(树形dp)

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题目链接：http://poj.org/problem?id=3140

思路：简单树形dp题，dp[u]表示以u为根的子树的人数和。

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<climits> 7 #include<algorithm> 8 #include<stack> 9 #include<map>10 #include<set>11 #include<vector>12 #include<queue>13 #include<list>14 using namespace std;15 #define MAXN 11111116 #define inf 1<<3017 #define INF 1LL<<6018 typedef long long ll;19 typedef pair<int,int>PP;20 template<class T> inline T Get_MIN(const T &a,const T &b){ return a < b ? a : b; }21 template<class T> inline T Get_MAX(const T &a,const T &b){ return a > b ? a : b; }22 template<class T> inline T ABS(const T &a){ return a < 0 ? -a : a; }23 24 struct Edge{25     int v,next;26 }edge[MAXN<<4];27 28 int n,m,NE;29 int head[MAXN];30 31 void Insert(int u,int v)32 {33     edge[NE].v=v;34     edge[NE].next=head[u];35     head[u]=NE++;36 }37 38 ll dp[MAXN],sum,MIN;39 void dfs(int u,int father)40 {41     for(int i=head[u];i!=-1;i=edge[i].next){42         int v=edge[i].v;43         if(v==father)continue;44         dfs(v,u);45         dp[u]+=dp[v];46     }47     MIN=Get_MIN(MIN,ABS((dp[u]-(sum-dp[u]))));48    // cout<<u<<"**"<<dp[u]<<"****"<<abs(sum-dp[u])<<"**"<<MIN<<endl;49 }50 51 52 int main()53 {54     int u,v,t=1;55     while(~scanf("%d%d",&n,&m)&&(n||m)){56         NE=0;57         memset(head,-1,sizeof(head));58         sum=0;59         for(int i=1;i<=n;i++)scanf("%lld",&dp[i]),sum+=dp[i];60         while(m--){61             scanf("%d%d",&u,&v);62             Insert(u,v);63             Insert(v,u);64         }65         MIN=INF;66         dfs(1,-1);67         printf("Case %d: %lld\n",t++,MIN);68     }69     return 0;70 }

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