MUTC 1 B - Hourai Jeweled 树形dp?

Hourai Jeweled

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 163840/163840 K (Java/Others)
Total Submission(s): 810    Accepted Submission(s): 291

Problem Description

Kaguya Houraisan was once a princess of the Lunarians, a race of people living on the Moon. She was exiled to Earth over a thousand years ago for the crime of using the forbidden Hourai Elixir to make herself immortal. Tales of her unearthly beauty led men from all across the land to seek her hand in marriage, but none could successfully complete her trial of the Five Impossible Requests. 

One of these requests is to reckon the value of "Hourai Jeweled (蓬莱の玉の枝)". The only one real treasure Kaguya has, in her possession. As showed in the picture, Hourai Jeweled is a tree-shaped twig. In which, each node is ornamented with a valuable diamond and each edge is painted with a briliant color (only bright man can distinguish the difference). Due to lunarians' eccentric taste, the value of this treasure is calculated as all the gorgeous roads' value it has. The road between two different nodes is said to be gorgeous, if and only if all the adjacent edges in this road has diffenrent color. And the value of this road is the sum of all the nodes' through the road.
Given the value of each node and the color of each edge. Could you tell Kaguya the value of her Hourai Jeweled?

 

Input

The input consists of several test cases. 
The first line of each case contains one integer N (1 <= N <= 300000), which is the number of nodes in Hourai Jeweled.
The second line contains N integers, the i-th of which is Vi (1 <= Vi <= 100000), the value of node i.
Each of the next N-1 lines contains three space-separated integer X, Y and Z (1<=X,Y<=N, 1 <= Z <= 100000), which represent that there is an edge between X and Y painted with colour Z. 

 

Output

For each test case, output a line containing the value of Hourai Jeweled.

 

Sample Input

   
   
   
   

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

 

Sample Output

   
   
   
   

    
    
    
    134

    
    
    
    
 
     
 
      Hint 
     
 gorgeous roads are : 1-2 Value: 8 1-3 Value: 9 1-4 Value:13 1-2-6 Value:12 2-1-3 Value:11 2-1-4 Value:15 2-5 Value:3 2-6 Value:6 3-1-4 Value:16 3-1-2-6 Value:15 4-1-2-6 Value:19 5-2-6 Value:7 
    
    
    
    
     
   
   
   
   

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

Recommend

zhuyuanchen520

 

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转题解:

一次dfs就能搞定的问题，其实没什么复杂的算法，主要难点在于统计时的一些小技巧，个人难度评价是偏难的中档题。
从任意一点开始深搜，每颗子树搜索完毕之后向上返回pair<可以延伸到该点且最后一条边与由父节点到该点的边颜色不同的gorgeous边的条数 , 所有这种边分数的总和>
每次深搜完一个子节点之后，增加的过这一点的gorgeous边的总分数为：
    之前深搜的所有子节点向上返回的边数之和 * 当前子节点返回的分数 +
    之前深搜的所有子节点向上返回的分数之和 * 当前子节点返回的边数 +
    之前深搜的所有子节点向上返回的边数之和 * 当前子节点返回的边数 * 当前点的权

用O(n)的时间把所有的累计起来就好了什么的

才没有呢 = =
如果有当前节点分别到两个子节点的边的颜色相同的话也是不行的,
由于数据中添加了度数很大的点，理论上O(儿子数^2)的两两统计也是要被卡掉的 （希望确实的做到了
正确的做法是对所有的儿子按边的颜色排个序，然后在按这个序深搜和统计

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对照着标程敲了一遍orz

一定是我树形dp的姿势不对

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

const int maxn=1111111;

__int64 ans,key[maxn],num[maxn],cnt[maxn];
int n;
vector< pair<int,int> >tree[maxn];

void dfs(int v,int pa,int pac)
{
    __int64 lastnum=0,nownum=0,lastcnt=0,nowcnt=0;
    int lastcolor=-1;
    num[v]=key[v];
    cnt[v]=1;
    for (int i=0;i<tree[v].size();i++)
    {
        if (tree[v][i].second!=pa)
        {
            dfs(tree[v][i].second,v,tree[v][i].first);
            if (tree[v][i].first!=lastcolor)
            {
                lastnum=nownum;
                lastcnt=nowcnt;
            }
            lastcolor=tree[v][i].first;
            ans+=lastnum*cnt[tree[v][i].second];
            ans+=lastcnt*num[tree[v][i].second];
            ans+=lastcnt*cnt[tree[v][i].second]*key[v];
            nownum+=num[tree[v][i].second];
            nowcnt+=cnt[tree[v][i].second];
            if (pac!=tree[v][i].first)
            {
                cnt[v]+=cnt[tree[v][i].second];
                num[v]+=num[tree[v][i].second];
                num[v]+=cnt[tree[v][i].second]*key[v];
            }
        }
    }
    if (v!=1)
    {
        ans+=cnt[v]*key[pa];
        ans+=num[v];
    }
}

int main()
{
    while (~scanf("%d",&n))
    {
        for (int i=1;i<=n;i++)
        {
            scanf("%I64d",&key[i]);
            tree[i].clear();
        }
        for (int i=1;i<=n-1;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            tree[a].push_back(make_pair(c,b));
            tree[b].push_back(make_pair(c,a));
        }
        for (int i=1;i<=n;i++)
        {
            sort(tree[i].begin(),tree[i].end());
        }
        ans=0;
        dfs(1,0,-1);
        printf("%I64d\n",ans);
    }
    return 0;
}