The more, The Better

Time Limit: 6000/2000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5414????Accepted Submission(s): 3217


Problem Description ACboy非常喜歡玩一種戰(zhàn)略游戲，在一個地圖上，有N座城堡。每座城堡都有一定的寶物，在每次游戲中ACboy同意攻克M個城堡并獲得里面的寶物。但因為地理位置原因。有些城堡不能直接攻克，要攻克這些城堡必須先攻克其它某一個特定的城堡。你能幫ACboy算出要獲得盡量多的寶物應(yīng)該攻克哪M個城堡嗎？

Input 每一個測試實例首先包含2個整數(shù)，N,M.(1 <= M <= N <= 200);在接下來的N行里。每行包含2個整數(shù)。a,b. 在第 i 行，a 代表要攻克第 i 個城堡必須先攻克第 a 個城堡，假設(shè) a = 0 則代表能夠直接攻克第 i 個城堡。

b 代表第 i 個城堡的寶物數(shù)量, b >= 0。當N = 0, M = 0輸入結(jié)束。

Output 對于每一個測試實例。輸出一個整數(shù)。代表ACboy攻克M個城堡所獲得的最多寶物的數(shù)量。
Sample Input 3 2 0 1 0 2 0 3 7 4 2 2 0 1 0 4 2 1 7 1 7 6 2 2 0 0
Sample Output 5 13

dp[i][j]表示以i為根節(jié)點j個子節(jié)點的最大值。

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> #include<vector> typedef long long LL; using namespace std; const int maxn=220; int v[maxn]; int n,m; int dp[maxn][maxn]; vector<int>s[maxn]; void tree_dp(int n,int f) {int len=s[n].size();dp[n][1]=v[n];for(int i=0;i<len;i++){if(f>1) tree_dp(s[n][i],f-1);for(int j=f;j>=1;j--){for(int k=1;k<=j;k++)dp[n][j+1]=max(dp[n][j+1],dp[n][j+1-k]+dp[s[n][i]][k]);}} } int main() {int f;while(~scanf("%d%d",&n,&m)&&(n+m)){v[0]=0;memset(dp,0,sizeof(dp));for(int i=0;i<=n;i++)s[i].clear();for(int i=1;i<=n;i++){scanf("%d%d",&f,&v[i]);s[f].push_back(i);}tree_dp(0,m+1);printf("%d\n",dp[0][m+1]);}return 0; }

Anniversary party

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?4329	?	Accepted:?2463

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 N – 1 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

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> typedef long long LL; using namespace std; const int maxn=6005; int dp[maxn][2],pre[maxn]; int visit[maxn],n; void tree_dp(int x) {visit[x]=1;for(int i=1;i<=n;i++){ // cout<<"111 "<<i<<endl;if(!visit[i]&&pre[i]==x){tree_dp(i);dp[x][1]+=dp[i][0];dp[x][0]+=max(dp[i][1],dp[i][0]);}} }int main() {while(~scanf("%d",&n)){memset(dp,0,sizeof(dp));memset(visit,0,sizeof(visit));memset(pre,0,sizeof(pre));for(int i=1;i<=n;i++)scanf("%d",&dp[i][1]);int x,y,root;while(~scanf("%d%d",&x,&y)&&(x+y)){pre[x]=y;root=y;}while(pre[root])root=pre[root];// cout<<"fuck "<<root<<endl;tree_dp(root);printf("%d\n",max(dp[root][0],dp[root][1]));}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/yxwkf/p/4605279.html

總結(jié)

以上是生活随笔為你收集整理的DP：树DP的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。