POJ 3140 Contestants Division (水题)

思路：记录每个节点的子节点的总和和非子节点的总和的差的绝对值，再扫一遍就完了。水题。。。和树形DP没关系。。。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

#define abs(x)  ((x) >= 0 ? (x) : -(x))
using namespace std;
typedef __int64 LL;
const LL maxn = 100005;
const LL INF = 0x3f3f3f3f3f3f3f3f;

struct Nod{
    LL b,next;
    void init(LL b,LL next){
        this->b=b;this->next=next;
    }
}buf[maxn<<1];
LL len,E[maxn];

LL n,m,sum,vi[maxn],num[maxn],dp[maxn];

void init(){
    len=0;sum=0;
    memset(E,-1,sizeof(E));
    memset(dp,0,sizeof(dp));
}

void add_edge(LL a,LL b){
    buf[len].init(b,E[a]);E[a]=len++;
    buf[len].init(a,E[b]);E[b]=len++;
}

void dfs(LL u,LL pre){
    LL i,v;
    num[u] = vi[u];
    for(i=E[u];i!=-1;i=buf[i].next){
        v=buf[i].b;
        if(v == pre) continue;
        dfs(v,u);
        num[u] += num[v];
    }
    dp[u] = abs(sum - 2*num[u]);
}

int main(){
    LL i,j,a,b,o,cas=0;
    while(scanf("%I64d%I64d",&n,&m),n+m){
        init();
        for(i=1;i<=n;i++){
            scanf("%I64d",&vi[i]);
            sum += vi[i];
        }
        while(m--){
            scanf("%I64d%I64d",&a,&b);
            add_edge(a,b);
        }
        dfs(1,-1);
        o = INF;
        //cout<<sum<<endl;
        //for(int i=1;i<=n;i++) cout<<num[i]<<" "; cout<<endl;
        //for(int i=1;i<=n;i++) cout<<dp[i]<<" "; cout<<endl;
        for(int i=1;i<=n;i++) o = min(o,dp[i]);
        printf("Case %I64d: %I64d\n",++cas,o);
    }
    return 0;
}