1380 Prom without a boss

1380 Prom without a boss

time limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingTitle Description Description

Ural University has a staff of N and is numbered 1~n. They have affiliation, which means that their relationship is like a tree rooted in the headmaster, and the parent node is the direct boss of the child node. Each employee has a happiness index. There is now an anniversary party, which asks the staff to have the most happiness index. However, no staff is willing to attend with the direct boss.

Enter a description Input Description

The first line is an integer n. (1<=n<=6000)
Next n lines, line i+1 represents the happiness index RI for staff i. ( -128<=ri<=127)
Next N-1 line, enter a pair of integer l, K for each line. Indicates that K is the direct boss of L.
Last line input 0, 0.

Output description Output Description

Output the maximum happiness index.

Sample input 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 Sample Output

5

Data range and Tips Data Size & Hint

Each test point 1s

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Topic simplified version (same idea)

Example code

1#include <cstdio>2#include <iostream>3#include <vector>4 #defineN 550005 using namespacestd;6 intf[n][2],fa[n],n;7vector<int>V[n];8 voiddpintx)9 {Ten      for(intj=0; J<v[x].size (); j + +){ One         intI=V[x][j]; A DP (i); -f[x][0]+=max (f[i][0],f[i][1]); -f[x][1]+=f[i][0]; the     } -f[x][1]++; - } - intMain () + { -scanf"%d",&n); +      for(intI=1, x,y;i<n;i++){ Ascanf"%d%d",&x,&y); atfa[x]=y; - v[y].push_back (x); -     } -      for(intI=1; i<=n;i++) -         if(fa[i]==0){ - DP (i); inprintf"%d\n", Max (f[i][0],f[i][1])); -             return 0; to         } +}

is the tree-type DP Board

1380 Prom code without a boss

1#include <cstdio>2#include <iostream>3#include <vector>4 #defineN 550005 using namespacestd;6 intf[n][2],fa[n],a[n],n,x,y;7vector<int>V[n];8 voiddpintx)9 {Ten      for(intj=0; J<v[x].size (); j + +) {//how many edges of the loop are connected to x One         intI=V[X][J];//One of the sons of X ADP (i);//recursion to the bottom -f[x][0]+=max (f[i][0],f[i][1]);//if x is not selected, the lower level can be selected or not selected -f[x][1]+=f[i][0];//if X is selected, the lower layer must not be selected the     } -f[x][1]+=A[X];//if X is selected, add itself - } - intMain () + { -scanf"%d",&n); +      for(intI=1; i<=n;i++) Ascanf"%d", A +i);  at      while(SCANF ("%d%d", &x,&y) = =2&&x&&y) { -Fa[x]=y;//X's father is Y. -V[y].push_back (x);//V[y (father number)][0-number of sides (table)]=x (son number); -     } -      for(intI=1; i<=n;i++) -         if(fa[i]==0){//Find the root in DP (i); -printf"%d\n", Max (f[i][0],f[i][1]));//Root may have more than one son, choose not to choose the best root to             return 0; +         } -}

1380 Prom without a boss