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  • BNUOJ 2461 Anniversary party

    Anniversary party

    Time Limit: 1000ms
    Memory Limit: 65536KB
    This problem will be judged on PKU. Original ID: 2342
    64-bit integer IO format: %lld      Java class name: Main
     
     
    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

    Source

     
    解题:。。。。
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define pii pair<int,int>
    15 #define INF 0x3f3f3f3f
    16 using namespace std;
    17 const int maxn = 6010;
    18 int dp[maxn][2],w[maxn],n,in[maxn];
    19 vector<int>g[maxn];
    20 void dfs(int u,int fa){
    21     dp[u][1] = w[u];
    22     for(int i = 0; i < g[u].size(); i++){
    23         if(g[u][i] == fa) continue;
    24         dfs(g[u][i],u);
    25         dp[u][1] += dp[g[u][i]][0];//u来,g[u][i]不来
    26         dp[u][0] += max(dp[g[u][i]][1],dp[g[u][i]][0]);
    27         //u不来,g[u][i]来或者不来
    28     }
    29 }
    30 int main() {
    31     int i,j,u,v,tmp;
    32     while(~scanf("%d",&n)){
    33         for(i = 1; i <= n; i++){
    34             scanf("%d",w+i);
    35             g[i].clear();
    36             in[i] = 0;
    37             dp[i][0] = dp[i][1] = 0;
    38         }
    39         for(i = 1; i; i++){
    40             scanf("%d %d",&u,&v);
    41             if(u+v == 0) break;
    42             g[u].push_back(v);
    43             g[v].push_back(u);
    44             in[v]++;
    45         }
    46         for(i = 1; i <= n; i++){
    47             if(!in[i]){dfs(i,-1);break;}
    48         }
    49         printf("%d
    ",max(dp[i][0],dp[i][1]));
    50     }
    51     return 0;
    52 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3916527.html
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