【HDOJ 5834】Magic boy Bi Luo with his excited tree（树型DP）

【HDOJ 5834】Magic boy Bi Luo with his excited tree（树型DP）

Magic boy Bi Luo with his excited tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem 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 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
    
    
    
    

 

Author

UESTC

 

Source

2016中国大学生程序设计竞赛 - 网络选拔赛

题目大意：
一棵树，节点有价值，树边有花费。
节点价值只能get到一次，树边花费每次经过都要消耗。
输出从节点1~n出发，分别能得到的最大净价值(节点收获-树边花费)

一眼树型dp

可能是做树dp做多了有感觉了，或者是因为训练计划上类似的题较多，。

首先考虑子问题：
节点1做根，输出最大价值。
令 dp[u][0] 表示从u出发，往下遍历，返回u的最大价值。
dp[u][1] 表示从u出发，往下遍历，不回u的最大价值。
这样转移其实就是
dp[u][1]=max(dp[u][0]+dp[v][1]+wuv,dp[v][0]+dp[u][1]+2⋅wuv)
dp[u][0]=∑dp[v][0]+2⋅wuv

回到此题，需要求得其实是以1 ~ n为根的最大价值。
那么第一遍dfs求出上面的数组。
第二遍dfs当从u遍历v时，给v传递两个参数：
从u出发，不经过v，回到u的最大价值
从u出发，不经过v，不回u的最大价值

这样当遍历v时，考虑与 dp[v][0] 还有 dp[v][1] 组合一下，去个最大即可。
求不经过v回到u的最大价值好办，第一遍dfs的时候对路标记一下，如果经过了v，去掉v这条路上获得的最大价值即可。
但是求不经过v，不回到u的最大价值时，如果 dp[u][1] 经过v了，单单除去v这条路，又变成了回到u的最大价值。
因此考虑给dp再加一个记录，对于u而言，第一遍dfs找到回到u的最大价值和不会到u的最大以及次大价值。

代码如下：

#include 

using namespace std;
const int mxn = 112345;

struct Edge
{
    int v,w,next,f;
};

Edge eg[mxn*2];
int head[mxn],val[mxn];
int ans[mxn];
int dp[mxn][3],nxt[mxn][3],tp;

void Add(int u,int v,int w)
{
    eg[tp].v = v;
    eg[tp].w = w;
    eg[tp].next = head[u];
    eg[tp].f = 0;
    head[u] = tp++;
}

void dfs1(int u,int pre)
{
    dp[u][0] = dp[u][1] = dp[u][2] = val[u];
    nxt[u][0] = nxt[u][1] = nxt[u][2] = -1;
    int v,w;
    for(int i = head[u]; i != -1; i = eg[i].next)
    {
        v = eg[i].v;
        if(v == pre) continue;
        w = eg[i].w;
        dfs1(v,u);

        dp[u][1] += max(0,dp[v][0]-2*w);
        dp[u][2] += max(0,dp[v][0]-2*w);

        if(w <= dp[v][1])
        {
            int tmp = dp[u][0]+dp[v][1]-w;

            if(tmp > dp[u][1])
            {
                if(dp[u][1] > dp[u][2])
                {
                    dp[u][2] = dp[u][1];
                    nxt[u][2] = nxt[u][1];
                }
                dp[u][1] = tmp;
                nxt[u][1] = i;
            }
            else if(tmp > dp[u][2])
            {
                dp[u][2] = tmp;
                nxt[u][2] = i;
            }
        }

        if(w*2 <= dp[v][0])
        {
            eg[i].f = 1;
            dp[u][0] += dp[v][0]-w*2;
        }
    }
}

void dfs2(int u,int pre,int bk,int nbk)
{
    bk = max(bk,0);
    nbk = max(nbk,0);
    //printf("u = %d bk = %d nbk = %d\n",u,bk,nbk);
    int v,w;
    ans[u] = max(dp[u][0]+nbk,dp[u][1]+bk);

    int nbk1,nbk2,bbk;
    for(int i = head[u]; i != -1; i = eg[i].next)
    {
        v = eg[i].v;
        w = eg[i].w;
        if(v == pre) continue;

        bbk = dp[u][0];
        nbk1 = dp[u][1];
        nbk2 = dp[u][2];

        if(eg[i].f)
        {
            bbk = bbk-dp[v][0]+w*2;
            nbk1 = nbk1-dp[v][0]+2*w;
            nbk2 = nbk2-dp[v][0]+2*w;
        }
        if(nxt[u][1] == i)
        {
            nbk1 = dp[u][1]-dp[v][1]+w;
        }
        if(nxt[u][2] == i)
        {
            nbk2 = dp[u][2]-dp[v][1]+w;
        }

        dfs2(v,u,bbk-2*w+bk,max(bbk+nbk-w,bk+max(nbk1,nbk2)-w));
    }
}

int main()
{
    int t,u,v,w,n;

    scanf("%d",&t);

    for(int z = 1; z <= t; ++z)
    {
        printf("Case #%d:\n",z);

        memset(head,-1,sizeof(head));
        tp = 0;

        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) scanf("%d",&val[i]);
        for(int i = 1; i < n; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            Add(u,v,w);
            Add(v,u,w);
        }

        dfs1(1,1);
        dfs2(1,1,0,0);

        for(int i = 1; i <= n; ++i)
        {
//            printf("%d:\n",i);
//            printf("back:%d\n",dp[i][0]);
//            printf("nbk1:%d %d\n",nxt[i][1] == -1? -1: eg[nxt[i][1]].v,dp[i][1]);
//            printf("nbk2:%d %d\n",nxt[i][2] == -1? -1: eg[nxt[i][2]].v,dp[i][2]);
            printf("%d\n",ans[i]);
        }
    }

    return 0;
}