HDU 4303 Hourai Jeweled 解题报告

HDU 4303 Hourai Jeweled 解题报告

评测地址： http://acm.hdu.edu.cn/showproblem.php?pid=4303

评测地址： https://xoj.red/contests/view/1155/1

题目描述

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?

输入格式

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.

输出格式

Output

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

样例输入输出

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

样例解释

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

版权信息

Author

BUPT

Source

2012 Multi-University Training Contest 1

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题目翻译

题目描述：

一棵N个点的树，树上每个点有点权，每条边有颜色。一条路径的权值是这条路径上所有点的点权和，一条合法的路径需要满足该路径上任意相邻的两条边颜色都不相同。问这棵树上所有合法路径的权值和是多少啊？（无向路径）

输入格式：

第一行一个整数N，代表树上有多少个点。

接下来一行N个整数，代表树上每个点的权值。

接下来N-1行，每行三个整数S、E、C，代表S与E之间有一条颜色为C的边

输出格式：

一行一个整数，代表所求的值。

数据范围：

对与30%的数据，1≤N≤1000。

对于另外20%的数据，可用的颜色数不超过1e9且随机数据。

对于另外20%的数据，树的形态为一条链。

对于100%的数据，1≤N≤3e5，可用的颜色数不超过1e9，所有点权的大小不超过1e5。

题目解析（sol）

解析

l  Sub1 对于30%的数据，1≤N≤1000。

暴力dfs枚举所有链

复杂度： O(n^2)

程序：见后附1

l  Sub2 对于另外20%的数据，树的形态为一条链。

线性dp。

设f[i]表示到第i位的路径总长度，显然singma{f[i]|1<=i<=n}就是答案

另外设cnt[i]表示第i个点计入几次。

初始值：f[1]=0，由题意可知一个点不算

转移：

若上一条链和这一条链颜色一样，

cnt[i]=1，就是说前面的都不可以走到这个点上。

f[i]=w[r[i-1]]+w[r[i]];

            若上一条链和这一条颜色不一样，

              cnt[i]=cnt[i-1]+1，前面的点都可以走到这个点上

              f[i]=f[i-1]+cnt[i]*w[r[i]]+w[r[i-1]]

    复杂度：对于链的数据O(n)，对于树的数据O(n^2)

程序：（见后附1）

l  Sub3 100%树形dp+统计

考虑两种可能：1.儿子到父亲的一条链

2.儿子跨过父亲到儿子

对于每个节点我们记录两个值就是：

1.这个节点可用子路径数量num[]

2.以节点为子树根向下链所有路径总长度dp[]

   在考虑儿子跨过父亲到儿子的情况下我们发现暴力枚举每一个儿子是会T的，想到用到mpnum[col]表示和父节点相邻边为col色的路径条数，mpdp表示子节点和父亲连边为col色的点权和。

  把两种情况分开来考虑。

1. 儿子到父亲的一条链:

Dp[u]=Dp[u]+Dp[v]+val[u]*num[v] （lastcolor != nowcolor）

Dp[u]=Dp[u] （lastcolor == nowcolor）

1. 儿子跨过父亲到儿子

其中s_Dp表示子树的Dp[u]的和，Num[u]表示子树总的经过次数,Ans是答案

Temp+=Dp[v]*(sum_Num-mpnum[a[i].col])+Num[v]*(S_Dp-mpdp[a[i].col])

Ans+=temp/2;

程序:（见后附2）

程序（Sub1+Sub2 ： 50pts）

# include <bits/stdc++.h>
# define int long long
using namespace std;
const int MAXN=3e5 + 10;
int w[MAXN],head[MAXN],n,ans=0,tot=0,tt=0,r[MAXN],du[MAXN],f[MAXN];
int cnt[MAXN];
bool vis[MAXN];
struct rec{
    int pre,to,col;
}a[MAXN];
struct cc{
    int from,to,val,id;
}node[MAXN];
map<pair<int,int>,int>mp;
void adde(int u,int v,int col)
{
    a[++tot].pre=head[u];
    a[tot].to=v;
    a[tot].col=col;
    head[u]=tot; 
} 
void dfs(int u,int co,int sum)
{
    vis[u]=true;  sum+=w[u];  ans+=sum; 
    for (int i=head[u];i;i=a[i].pre) {
        int v=a[i].to; if (vis[v]) continue;
        if (co==a[i].col) continue;
        dfs(v,a[i].col,sum);
    }
}
void dfs2(int u)
{
    r[++tt]=u; vis[u]=true;
    for (int i=head[u];i;i=a[i].pre){
        int v=a[i].to; if (vis[v]) continue;
        dfs2(v);
    }
}
void work()
{
    memset(du,0,sizeof(du));
    for (int i=1;i<=n;i++) scanf("%lld",&w[i]);
    int u,v,col;
    for (int i=1;i<=n-1;i++) {
        scanf("%lld%lld%lld",&u,&v,&col);
        adde(u,v,col); adde(v,u,col);
        mp[make_pair(u,v)]=col;
        mp[make_pair(v,u)]=col;
        du[u]++,du[v]++;
    }
    int root=1;
    for (int i=1;i<=n;i++)
     if (du[i]==1) { root=i; break;}
    tt=0;  memset(vis,false,sizeof(vis));
    dfs2(root); 
    f[1]=0; int lc=-1;
    for (int i=2;i<=n;i++){
        if (mp[make_pair(r[i-1],r[i])]!=lc) {
            cnt[i]=cnt[i-1]+1;
            f[i]=f[i-1]+w[r[i]]*cnt[i]+w[r[i-1]];
        }
        else cnt[i]=1,f[i]=w[r[i]]+w[r[i-1]];
        lc=mp[make_pair(r[i-1],r[i])];
    }
    int ans=0;
    for (int i=1;i<=n;i++) ans+=f[i];
    printf("%lld
",ans);
}
signed main()
{
    scanf("%lld",&n);
    if (n>1000) { work(); return 0;}
    int t=0;
    for (int i=1;i<=n;i++) scanf("%lld",&w[i]),t+=w[i];
    int u,v,col;
    for (int i=1;i<=n-1;i++) {
        scanf("%lld%lld%lld",&u,&v,&col);
        adde(u,v,col); adde(v,u,col);
    }
    for (int i=1;i<=n;i++) {
        memset(vis,false,sizeof(vis));
        dfs(i,-1,0);
    } 
    ans=(ans-t)/2;
    printf("%lld
",ans);
    return 0;
}

程序（Sub3： 100 pts）

# include <bits/stdc++.h>
# define int long long
using namespace std;
const int MAXN=3e5+10;
int n,val[MAXN],ans;
struct rec{
    int pre,to,col;
}a[2*MAXN];
int head[MAXN],tot=0;
int num[MAXN],dp[MAXN];
void adde(int u,int v,int col)
{
    a[++tot].pre=head[u];
    a[tot].to=v;
    a[tot].col=col;
    head[u]=tot;
}
void dfs(int u,int fa,int lc)
{
    num[u]=1; dp[u]=val[u];
    map<int,int> mpnum,mpdp;
    int s_Num=0,s_Dp=0,t=0;
    for (int i=head[u];i;i=a[i].pre){
        int v=a[i].to; if(v==fa) continue;
        dfs(v,u,a[i].col);
        t=dp[v]+val[u]*num[v];
        if (a[i].col!=lc) {
            num[u]+=num[v];
            dp[u]+=t;
        }
        s_Num+=num[v];
        s_Dp+=t;
        mpnum[a[i].col]+=num[v];
        mpdp[a[i].col]+=t;
    }
    ans+=s_Dp;
    int temp=0;
    for (int i=head[u];i;i=a[i].pre) {
        int v=a[i].to; if (v==fa) continue;
        temp+=dp[v]*(s_Num-mpnum[a[i].col])+num[v]*(s_Dp-mpdp[a[i].col]);
    }
    ans+=temp/2;
}
signed main()
{
    scanf("%lld",&n);
    for (int i=1;i<=n;i++) scanf("%lld",&val[i]);
    int u,v,col;
    for (int i=1;i<=n-1;i++) {
        scanf("%lld%lld%lld",&u,&v,&col);
        adde(u,v,col);
        adde(v,u,col);
    }
    dfs(1,0,-1);
    printf("%lld
",ans); 
    return 0;
 }

From：     HGOI

Name：ljc20020730

Date： 20181004