URAL 1039. Anniversary Party

Author: 2015-10-31

ACM

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1. 题目

http://acm.timus.ru/problem.aspx?space=1&num=1039

1039. Anniversary Party

Time limit: 0.5 second
Memory limit: 8 MB

Background

The president of the Ural State University is going to make an 80’th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from 1 to N, The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.

Problem

Your task is to make up a guest list with the maximal conviviality rating of the guests.

Input

The first line of the input contains a number N. 1 ≤ N ≤ 6000. 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 the supervisor relation tree goes. Each line of the tree specification has the form
<L> <K>
which means that the K-th employee is an immediate supervisor of L-th employee. Input is ended with the line
0 0

Output

The output should contain the maximal total rating of the guests.

Sample

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

5
Problem Author: Marat Bakirov
Problem Source: Ural State University Internal Contest October’2000 Students Session

2. 思路

有编号为1~N的N名职员，给出每名职员的愉悦值，以及每名职员的直接上级。现要邀请这些职员参加一个愉悦的年会，不允许职员和其直接上级同时参加，求年会所能获得的最大愉悦值。

树形DP，类似于POJ 1463. Strategic game。上下级关系是一个树形的结构，每个职员的直接下级作为该职员的子节点，对于树中的任意节点s：

  • 使用maxRating[s][0]表示不在s不参加时，以s为根的子树上的人参加年会所能获取的最大愉悦值，此时s的所有直接下属e都可以参加(maxRating[e][1])，也可以不参加(maxRating[e][0])，取决于maxRating[e][1]和maxRating[e][0]那个较大；
  • 使用maxRating[s][1]表示不在s参加时，以s为根的子树上的人参加年会所能获取的最大愉悦值，此时s的所有直接下属e都不能参加（maxRating[e][0]）。

即：

maxRating[1 : N][0 : 1] = 0 
for each child e of s:
    maxRating[s][0] += max(maxRating[e][1], maxRating[e][0]);
    maxRating[s][1] += maxRating[e][0];

3. 代码

#include <cstdio>

#define MAX_N 6001

typedef struct EmployeeRecord {
    int id;
    struct EmployeeRecord *next;
} Employee;

void solvePE8c_AnniversaryParty();
int input(Employee *employees[], int rating[], int n);
void dfs(int id);
int max(int a, int b);

int main() {
    // freopen("test.txt", "r", stdin);
    solvePE8c_AnniversaryParty();
    return 0;
}

Employee *employees[MAX_N];
int rating[MAX_N], maxRating[MAX_N][2];
void solvePE8c_AnniversaryParty() {
    int n;
    scanf("%d", &n);
    int president = input(employees, rating, n);

    dfs(president);

    int result = max(maxRating[president][0], maxRating[president][1]);
    printf("%d\n", result);
}

void dfs(int id) {
    maxRating[id][0] = 0;
    maxRating[id][1] = rating[id];

    if (employees[id] != NULL) {
        maxRating[id][0] = 0;
        maxRating[id][1] = rating[id];
        Employee *e = employees[id];
        while (e != NULL) {
            dfs(e->id);
            maxRating[id][0] += max(maxRating[e->id][1], maxRating[e->id][0]);
            maxRating[id][1] += maxRating[e->id][0];
            e = e->next;
        }
    }
}

int max(int a, int b) {
    return a > b ? a : b;
}

bool isEmployee[MAX_N];
int input(Employee *employees[], int rating[], int n) {
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &rating[i]);
    }

    int employee, superv;
    scanf("%d %d", &employee, &superv);
    while (employee != 0) {
        Employee *e = new Employee();
        e->id = employee;
        e->next = employees[superv];
        employees[superv] = e;
        isEmployee[employee] = true;

        scanf("%d %d", &employee, &superv);
    }

    for (int i = 1; i <= n; ++i) {
        if (!isEmployee[i]) {
            return i;
        }
    }

    return -1;
}