HDU1520 Anniversary party 樹(shù)形動(dòng)態(tài)規(guī)劃

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15486????Accepted Submission(s): 5958

Problem Description

There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings.

Input

Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go T lines that describe a supervisor relation tree. Each line of the tree specification has the form:L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line 0 0

Output

Output should contain the maximal sum of guests' ratings.

Sample Input

7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0

Sample Output

5

Source

Ural State University Internal Contest October'2000 Students Session

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題意：

在一棵有根樹(shù)上每個(gè)節(jié)點(diǎn)有一個(gè)權(quán)值，相鄰的父親和孩子只能選擇一個(gè)，求總權(quán)值之和最大

思路：

這個(gè)DFS過(guò)程大家可能不太理解，dfs(son)貌似沒(méi)有寫(xiě)終止條件，可是卻沒(méi)引起無(wú)限遞歸，為什么呢？遞歸是怎么終止的？
題目大意是，在一棵有根樹(shù)上，每個(gè)節(jié)點(diǎn)有一個(gè)權(quán)值，相鄰的父親和孩子只能選擇一個(gè)，
求總權(quán)值之和最大.

首先，我們用a【i】存放節(jié)點(diǎn)i的權(quán)值，father【i】表示節(jié)點(diǎn)i的祖先，G【i】是一個(gè)vector動(dòng)態(tài)數(shù)組，存放i的所有子孫節(jié)點(diǎn)，

接下來(lái)我們尋找樹(shù)的根節(jié)點(diǎn)，從根節(jié)點(diǎn)開(kāi)始搜索，遍歷根節(jié)點(diǎn)的子孫節(jié)點(diǎn)，遞歸深度即數(shù)組G【root】的大小，

所以dfs(son)貌似沒(méi)有寫(xiě)終止條件，卻沒(méi)引起無(wú)限遞歸

dp[i][0]表示第i個(gè)人參加，dp[i][1]表示第i個(gè)人不參加?

狀態(tài)轉(zhuǎn)移方程：?

dp[root][0]+=max(dp[son][1],dp[son][0]);
當(dāng)前這個(gè)點(diǎn)不選，他的孩子可選、可不選，取最大值?
dp[root][1]+=dp[son][0];

當(dāng)前這個(gè)點(diǎn)選擇，它的孩子就不能選擇

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#include<bits/stdc++.h> using namespace std; const int maxn=6005; int father[maxn];//標(biāo)記根節(jié)點(diǎn) int n,a[maxn];//每個(gè)節(jié)點(diǎn)的權(quán)值 vector<int>G[maxn];//存放子節(jié)點(diǎn) int dp[maxn][2]; void dfs(int root) {dp[root][1]=a[root];for(int i=0;i<G[root].size();i++){int son=G[root][i];dfs(son);dp[root][0]+=max(dp[son][1],dp[son][0]);//當(dāng)前這個(gè)人不選，取選孩子和不選孩子的較大者dp[root][1]+=dp[son][0];//當(dāng)前這個(gè)人選，不選他的孩子} } int main() {ios::sync_with_stdio(0);int n;while(cin>>n){//memset(father,-1,sizeof(father));for(int i=1;i<=n;i++){cin>>a[i];G[i].clear();//清空dp[i][0]=dp[i][1]=0;father[i]=-1;}int l,k;while(cin>>l>>k,l+k){//l是k的孩子father[l]=k;//l的祖先為kG[k].push_back(l);}int root=1;while(father[root]!=-1)//尋找根節(jié)點(diǎn)root=father[root];dfs(root);cout<<max(dp[root][1],dp[root][0])<<endl;}return 0; }

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