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2013
12-12

HDU 1520 Anniversary party-动态规划-[解题报告] C++

Anniversary party

问题描述 :

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.

输入:

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

输出:

Output should contain the maximal sum of guests’ ratings.

样例输入:

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

样例输出:

5

本文出自   http://blog.csdn.net/shuangde800

题目点击打开链接

题意:

给出一棵树 每个节点有权值 要求父节点和子节点不能同时取 求能够取得的最大值

思路:

树形dp的入门题

f[u][0]表示以u为顶点的子树,不选u点的情况下最大值
f[u][1]表示以u为顶点的子树,选u点的情况下最大值
那么,

f[u][0] = sum{ max{f[v][0], f[v][1]}, v是u的儿子节点}; //当不选u点时,它的儿子节点可以不选也可以选
f[u][1] = val[u] + sum{f[v][0], v是u的儿子节点} //当选了u点时,它的儿子节点必须是不能选

代码:

/**==========================================
 *   This is a solution for ACM/ICPC problem
 *
 *   @author: shuangde
 *   @blog: blog.csdn.net/shuangde800
 *   @email: [email protected]
 *===========================================*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;

typedef long long int64;
const int INF = 0x3f3f3f3f;
const double PI  = acos(-1.0);

const int MAXN = 6010;

vector<int>adj[MAXN];
int indeg[MAXN];
int val[MAXN];
int f[MAXN][2];
int vis[MAXN];
int n, m;

void dfs(int u){
    vis[u] = true;
    f[u][0] = 0;
    f[u][1] = val[u];
    for(int i=0; i<adj[u].size(); ++i){
        int v = adj[u][i];
        if(vis[v]) continue;
        dfs(v);
        f[u][0] += max(f[v][1], f[v][0]);
        f[u][1] += f[v][0];
    }
}

int main(){

    while(~scanf("%d", &n) && n){

        for(int i=1; i<=n; ++i) adj[i].clear();

        for(int i=1; i<=n; ++i)
            scanf("%d", &val[i]);

        memset(indeg, 0, sizeof(indeg));

            int u, v;
        while(~scanf("%d%d", &v, &u) && v+u){
            adj[u].push_back(v);
            ++indeg[v];
        }

        memset(f, 0, sizeof(f));
        for(int i=1; i<=n; ++i)if(!indeg[i]){
            memset(vis, 0, sizeof(vis));
            dfs(i);
            printf("%d\n", max(f[i][0], f[i][1]));
            break;
        }
    }

    return 0;
}

解题报告转自:http://blog.csdn.net/shuangde800/article/details/9731725


  1. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。