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=-=抓住叶节点往上揪

Tree chain problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1752    Accepted Submission(s): 561

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1e5+88;
int fa[maxn],dep[maxn],size[maxn],pos[maxn],bl[maxn],head[maxn];
int sum[maxn];
int dp[maxn],su[maxn],n,m;
vector<int>G[maxn];
struct node{
   int to,next;
}edge[maxn<<1];
struct cst{
   int x,y,z;
}road[maxn];
int tot,sz;
void init(){
   tot=sz=0;
   memset(sum,0,sizeof(sum));
   memset(head,-1,sizeof(head));
   for(int i=1;i<=n;++i) G[i].clear();
}
void add(int u,int v) {
   edge[tot].to=v;
   edge[tot].next=head[u];
   head[u]=tot++;
}
void sadd(int u,int val) {
   for( ; u<=n;u+=u&(-u))
    sum[u]+=val;
}
int getsum(int u) {
    int ret=0;
   for(;u;u-=u&(-u))
    ret+=sum[u];
   return ret;
}
void dfs1(int x){
    size[x]=1;
    for(int i=head[x];i+1;i=edge[i].next){
        int v=edge[i].to;
        if(v==fa[x]) continue;
        fa[v]=x;
        dep[v]=dep[x]+1;
        dfs1(v);
        size[v]+=size[x];
    }
}
void dfs2(int x,int chain)
{
    bl[x]=chain;
    pos[x]=++sz;
    int k=0;
    for(int i=head[x];i+1;i=edge[i].next){
        int v=edge[i].to;
        if(dep[v]>dep[x]&&size[v]>size[k])
            k=v;
    }
    if(!k) return;
    dfs2(k,chain);
    for(int i=head[x];i+1;i=edge[i].next)
        if(dep[edge[i].to]>dep[x]&&edge[i].to!=k)
        dfs2(edge[i].to,edge[i].to);
}
int LCA(int x,int y){
   while(bl[x]!=bl[y]) {
    if(dep[bl[x]]<dep[bl[y]]) swap(x,y);
    x=fa[bl[x]];
   }
   if(pos[x]>pos[y]) swap(x,y);
   return x;
}
int query(int x,int y){
    int ret=0;
   while(bl[x]!=bl[y]){
    if(dep[bl[x]]<dep[bl[y]]) swap(x,y);
    ret+=getsum(pos[x])-getsum(pos[bl[x]]-1);
    x=fa[bl[x]];
   }
   if(dep[x]>dep[y]) swap(x,y);
   ret+=getsum(pos[y])-getsum(pos[x]-1);
   return ret;
}
void solve(int u){
   su[u]=0;
   for(int i=head[u];i+1;i=edge[i].next){
    int v=edge[i].to;
    if(v==fa[u]) continue;
    solve(v);
    su[u]+=dp[v];
   }
   dp[u]=su[u];
   for(int i=0;i<(int)G[u].size();++i){
    int v=G[u][i];
    dp[u]=max(dp[u],query(road[v].x,road[v].y)+su[u]+road[v].z);
   }
   sadd(pos[u],su[u]-dp[u]);
}
int main(){
   int t,u,v;
   for(scanf("%d",&t);t--;){
    scanf("%d%d",&n,&m);
    init();
    for(int i=1;i<n;++i){
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    dfs1(1);
    dfs2(1,1);
    for(int i=1;i<=m;++i)
    {
        scanf("%d%d%d",&road[i].x,&road[i].y,&road[i].z);
        G[LCA(road[i].x,road[i].y)].push_back(i);
    }
    solve(1);
    printf("%d\n",dp[1]);
   }
}

HDU5293 树链剖分+树形DP

标签：output   nes   between   continue   esc   label   represent   ble   eve   

原文地址：http://www.cnblogs.com/mfys/p/7086592.html