HDU 4303 Hourai Jeweled 解题报告（树状DP+统计）

Hourai Jeweled

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 163840/163840 K (Java/Others)
Total Submission(s): 868    Accepted Submission(s): 315

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

66 2 3 7 1 41 2 11 3 21 4 32 5 12 6 2

 

Sample Output

134
Hint
gorgeous roads are :1-2 Value: 81-3 Value: 91-4Value:131-2-6   Value:122-1-3         Value:112-1-4 Value:152-5Value:32-6 Value:63-1-4        Value:163-1-2-6 Value:154-1-2-6 Value:195-2-6Value:7
 

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

    解题报告：树型DP。

    值有两种。一种是以父节点到子节点的路径，一种是子节点通过父节点到另一子节点的值。

    分析，易知每个节点应该保存当前节点的可用子路径数量，当前节点及其可用子节点的值得和。

    要注意的是，直接枚举儿子是否两两不同色会超时。可以用map（哈希应该也可以）保存某一颜色的值得和，或者先对边的颜色进行排序并统计。

    使用map的代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;typedef long long LL;const int maxn=600010;int first[maxn/2],vv[maxn],ww[maxn],nxt[maxn];LL dp[maxn/2];int num[maxn/2],node[maxn/2];int n;LL res;int eNum;void addEdge(int a,int b,int w){    nxt[eNum]=first[a],vv[eNum]=b,ww[eNum]=w,first[a]=eNum++;    nxt[eNum]=first[b],vv[eNum]=a,ww[eNum]=w,first[b]=eNum++;}void init(){    res=0;    eNum=2;    memset(first,-1,sizeof(first));}void dfs(int u,int p,int c){    num[u]=1;    dp[u]=node[u];    map<int,LL> mpNum,mpDp;    LL tmpNum=0,tmpDp=0,temp=0;    for(int a=first[u];~a;a=nxt[a]) if(vv[a]!=p)    {        int v=vv[a];        int col=ww[a];        dfs(v,u,col);        temp=dp[v]+(LL)node[u]*num[v];        if(col!=c)        {            num[u]+=num[v];            dp[u]+=temp;        }        tmpNum+=num[v];        tmpDp+=temp;        mpNum[col]+=num[v];        mpDp[col]+=temp;    }    res+=tmpDp;    temp=0;    for(int a=first[u];~a;a=nxt[a]) if(vv[a]!=p)        temp+=dp[vv[a]]*(tmpNum-mpNum[ww[a]])+num[vv[a]]*(tmpDp-mpDp[ww[a]]);    res+=temp/2;}int main(){    while(~scanf("%d",&n))    {        init();        for(int i=1;i<=n;i++)            scanf("%d",node+i);        for(int i=1;i<n;i++)        {            int a,b,w;            scanf("%d%d%d",&a,&b,&w);            addEdge(a,b,w);        }        dfs(1,0,0);        printf("%I64d\n",res);    }}

    边排序的代码，注意指向父节点的边可能影响到连续相同颜色的统计。

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <vector>using namespace std;typedef long long LL;const int maxn=300010;LL dp[maxn];int num[maxn],node[maxn];int n;LL res;struct Edge{int v,c;Edge(int _v,int _c){v=_v;c=_c;}bool operator<(const Edge& cmp) const{return c<cmp.c;}};vector<Edge> edge[maxn];void addEdge(int a,int b,int w){edge[a].push_back(Edge(b,w));edge[b].push_back(Edge(a,w));}void init(){res=0;for(int i=0;i<=n;i++)edge[i].clear();}void dfs(int u,int p,int c){num[u]=1;dp[u]=node[u];if(edge[u].size()==1) return;LL sumDp=0,nowDp=0,nowTemp=0,temp=0;int nowNum=0,sumNum=0;sort(edge[u].begin(),edge[u].end());for(vector<Edge>::iterator it=edge[u].begin();it!=edge[u].end();){if(it->v==p) {it++;continue;}int col=it->c;for(;it!=edge[u].end() && it->c==col;it++) if(it->v!=p){int v=it->v;dfs(v,u,col);temp=dp[v]+(LL)node[u]*num[v];if(col!=c){num[u]+=num[v];dp[u]+=temp;}res+=temp;nowNum+=num[v];nowDp+=dp[v];nowTemp+=temp;}res+=sumNum*nowDp+sumDp*nowNum;sumNum+=nowNum;sumDp+=nowTemp;nowNum=0;nowDp=0;nowTemp=0;}}int main(){while(~scanf("%d",&n)){init();for(int i=1;i<=n;i++)scanf("%d",node+i);for(int i=1;i<n;i++){int a,b,w;scanf("%d%d%d",&a,&b,&w);addEdge(a,b,w);}dfs(1,0,0);printf("%I64d\n",res);}}