UVA 12223 Moving to Nuremberg(树形DP)
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关键词:树中选取最优点
题意:简化后的数学模型:
已知一棵含点权和边权的无根树,在树中选择一个点u使得:
value[i]是i点的权值,d[u][i]是u点到i点的距离
解法:
1.以1为根化为有根树
2.求u=1的sum值
3.寻找1的子节点的sum值和1节点的sum值的关系,发现与子节点的权值和有关
拓展:
1.求
设w[i]是第i条边的权值,该边的尾节点是v,sum[v]是以v为根的子树权值和,sumvalue是树的总权
上式=
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<set>#include<vector>#include<map>#define ll long longusing namespace std;const int maxn = 50000+10;int t;ll n,m;ll s[maxn],value[maxn];ll sum;ll dis[maxn],d[maxn];//dis[i]:i到根节点的距离 d[i]:以i节点作为live点的代价ll ans[maxn],cnt,tmp;//cnt,待确定点的个数 tmp,待求点的代价struct Edge{ ll to,w,next;}edge[maxn<<1];ll head[maxn],tot;void add(ll u,ll v,ll w){edge[tot].to=v,edge[tot].w=w,edge[tot].next=head[u],head[u]=tot++;}void dfs1(ll u,ll f){ s[u]=value[u]; for(ll i=head[u];i!=-1;i=edge[i].next){ ll v=edge[i].to,w=edge[i].w; if(v==f) continue; dis[v]=dis[u]+w; dfs1(v,u); s[u]+=s[v]; }}void dfs2(ll u,ll f){ for(ll i=head[u];i!=-1;i=edge[i].next){ ll v=edge[i].to,w=edge[i].w; if(v==f) continue; d[v]=d[u]-s[v]*w+(sum-s[v])*w; dfs2(v,u); }}int main(){ scanf("%d",&t); while(t--){ scanf("%lld",&n); memset(head,-1,sizeof(head)); tot=0; memset(value,0,sizeof(value)); memset(dis,0,sizeof(dis)); memset(d,0,sizeof(d)); sum=0; for(ll i=0;i<n-1;i++){ ll x,y,z; scanf("%lld%lld%lld",&x,&y,&z); add(x,y,z),add(y,x,z); } scanf("%lld",&m); for(ll i=1;i<=m;i++){ ll u,v; scanf("%lld%lld",&u,&v); value[u]=v; sum+=value[u]; } dfs1(1,0); //tmp:以1作为live点的代价 for(ll i=1;i<=n;i++){ d[1]=d[1]+dis[i]*value[i]; } dfs2(1,0); cnt=0,tmp=0; for(ll i=1;i<=n;i++){ if(i==1||d[i]<tmp){ cnt=0,tmp=d[i]; ans[cnt++]=i; } else if(d[i]==tmp){ ans[cnt++]=i; } } printf("%lld\n",tmp*2); for(ll i=0;i<cnt-1;i++){ printf("%lld ",ans[i]); } if(cnt>=1) printf("%lld\n",ans[cnt-1]); } return 0;}
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