HDU5834 Magic boy Bi Luo with his excited tree（树形DP）

题目

Source

http://acm.hdu.edu.cn/showproblem.php?pid=5834

Description

Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it's value is V[i], and for each edge, there is a cost C[i], which means every time you pass the edge i , you need to pay C[i].

You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times.

Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i.

Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him?

Input

First line is a positive integer T(T≤10⁴) , represents there are T test cases.

Four each test：

The first line contain an integer N(N≤10⁵).

The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤10⁴).

For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it's cost is c(1≤c≤10⁴).

You can assume that the sum of N will not exceed 10⁶.

Output

For the i-th test case , first output Case #i: in a single line , then output N lines , for the i-th line , output ans[i] in a single line.

Sample Input

1
5
4 1 7 7 7
1 2 6
1 3 1
2 4 8
3 5 2

Sample Output

Case #1:
15
10
14
9
15

 

分析

题目大概说给一棵树，点和边都有权值，经过一点可以加上该点的权值但最多只加一次，经过边会减去该边权值，问从各个点分别出发最多能获得多少权值。

 

这题是很裸的一种树形DP，做过类似HDU2196就知道怎么做了。两个DFS分别在O(n)处理出两种信息，各个结点往其为根的子树走的信息和各个结点往父亲走的信息，各个结点就能在O(1)合并这两个信息分别得出各个结点的最终信息。。

对于这题需要的状态是：

• dp_down[0/1][u]：u结点往其为根的子树走，并且不走回来/走回来，能得到的最大权值
• dp_up[0/1][u]：u结点往其父亲向上走，并且不走回来/走回来，能得到的最大权值

转移的话，想想就知道了。。dp_up[1][u]可以通过计算各个孩子信息的最大值和次大值求得，其他的比较简单，不过麻烦。。细节要注意，逻辑好考虑清楚。。比赛时我就没考虑好几个逻辑WA了，然后死活找不到错，还好队友A了。。

 

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 111111

struct Edge{
    int v,w,next;
}edge[MAXN<<1];
int NE,head[MAXN];
void addEdge(int u,int v,int w){
    edge[NE].v=v; edge[NE].w=w; edge[NE].next=head[u];
    head[u]=NE++;
}

int val[MAXN];
int d_down[2][MAXN],d_up[2][MAXN];

void dfs1(int u,int fa){
    d_down[0][u]=d_down[1][u]=val[u];
    for(int i=head[u]; i!=-1; i=edge[i].next){
        int v=edge[i].v;
        if(v==fa) continue;
        dfs1(v,u);
        if(d_down[0][v]-2*edge[i].w>0) d_down[0][u]+=d_down[0][v]-2*edge[i].w;
    }
    int mx=0;
    for(int i=head[u]; i!=-1; i=edge[i].next){
        int v=edge[i].v;
        if(v==fa) continue;
        if(d_down[0][v]-2*edge[i].w>0){
            mx=max(mx,(d_down[1][v]-edge[i].w)-(d_down[0][v]-2*edge[i].w));
        }else{
            mx=max(mx,d_down[1][v]-edge[i].w);
        }
    }
    d_down[1][u]=d_down[0][u]+mx;
}
void dfs2(int u,int fa){

    int mx1=0,mx2=0,tmp;
    for(int i=head[u]; i!=-1; i=edge[i].next){
        int v=edge[i].v;
        if(v==fa) continue;
        if(d_down[0][v]-2*edge[i].w>0) tmp=(d_down[1][v]-edge[i].w)-(d_down[0][v]-2*edge[i].w);
        else tmp=d_down[1][v]-edge[i].w;
        if(mx1<tmp){
            mx2=mx1;
            mx1=tmp;
        }else if(mx2<tmp){
            mx2=tmp;
        }
    }

    for(int i=head[u]; i!=-1; i=edge[i].next){
        int v=edge[i].v;
        if(v==fa) continue;

        int tmp2;
        if(d_down[0][v]-2*edge[i].w>0){
            tmp2=d_down[0][u]-(d_down[0][v]-2*edge[i].w);
        }else{
            tmp2=d_down[0][u];
        }
        int mx=max(d_up[0][u]-2*edge[i].w,tmp2-2*edge[i].w);
        mx=max(mx,d_up[0][u]+tmp2-2*edge[i].w-val[u]);
        d_up[0][v]=val[v]+max(0,mx);

        if(d_down[0][v]-2*edge[i].w>0){
            if(mx1==(d_down[1][v]-edge[i].w)-(d_down[0][v]-2*edge[i].w)) tmp=d_down[1][u]-(d_down[1][v]-edge[i].w)+mx2;
            else tmp=d_down[1][u]-(d_down[0][v]-2*edge[i].w);
        }else if(d_down[1][v]-edge[i].w>0){
            if(mx1==d_down[1][v]-edge[i].w) tmp=d_down[1][u]-(d_down[1][v]-edge[i].w)+mx2;
            else tmp=d_down[1][u];
        }else tmp=d_down[1][u];
        mx=max(d_up[1][u]-edge[i].w,tmp-edge[i].w);
        mx=max(mx,max(d_up[0][u]+tmp-edge[i].w-val[u],d_up[1][u]+tmp2-edge[i].w-val[u]));
        d_up[1][v]=val[v]+max(0,mx);

        dfs2(v,u);
    }
}

int main(){
    int t;
    scanf("%d",&t);
    for(int cse=1; cse<=t; ++cse){
        NE=0;
        memset(head,-1,sizeof(head));
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; ++i){
            scanf("%d",val+i);
        }
        int a,b,c;
        for(int i=1; i<n; ++i){
            scanf("%d%d%d",&a,&b,&c);
            addEdge(a,b,c);
            addEdge(b,a,c);
        }
        dfs1(1,1);
        d_up[0][1]=d_up[1][1]=val[1];
        dfs2(1,1);
        printf("Case #%d:\n",cse);
        for(int i=1; i<=n; ++i){
            printf("%d\n",max(d_up[0][i]+d_down[1][i],d_up[1][i]+d_down[0][i])-val[i]);
        }
    }
    return 0;
}