HDU-1011 Starship Troopers(树形dp)
Starship Troopers
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15065 Accepted Submission(s): 4046
Problem Description You, the leader of Starship Troopers, are sent to destroy a base of the bugs. The base is built underground. It is actually a huge cavern, which consists of many rooms connected with tunnels. Each room is occupied by some bugs, and their brains hide in some of the rooms. Scientists have just developed a new weapon and want to experiment it on some brains. Your task is to destroy the whole base, and capture as many brains as possible.
To kill all the bugs is always easier than to capture their brains. A map is drawn for you, with all the rooms marked by the amount of bugs inside, and the possibility of containing a brain. The cavern’s structure is like a tree in such a way that there is one unique path leading to each room from the entrance. To finish the battle as soon as possible, you do not want to wait for the troopers to clear a room before advancing to the next one, instead you have to leave some troopers at each room passed to fight all the bugs inside. The troopers never re-enter a room where they have visited before.
A starship trooper can fight against 20 bugs. Since you do not have enough troopers, you can only take some of the rooms and let the nerve gas do the rest of the job. At the mean time, you should maximize the possibility of capturing a brain. To simplify the problem, just maximize the sum of all the possibilities of containing brains for the taken rooms. Making such a plan is a difficult job. You need the help of a computer.
Input The input contains several test cases. The first line of each test case contains two integers N (0 < N <= 100) and M (0 <= M <= 100), which are the number of rooms in the cavern and the number of starship troopers you have, respectively. The following N lines give the description of the rooms. Each line contains two non-negative integers – the amount of bugs inside and the possibility of containing a brain, respectively. The next N - 1 lines give the description of tunnels. Each tunnel is described by two integers, which are the indices of the two rooms it connects. Rooms are numbered from 1 and room 1 is the entrance to the cavern.
The last test case is followed by two -1’s.
Output For each test case, print on a single line the maximum sum of all the possibilities of containing brains for the taken rooms.
Sample Input 5 10 50 10 40 10 40 20 65 30 70 30 1 2 1 3 2 4 2 5 1 1 20 7 -1 -1
Sample Output 50 7
题意:就是一颗树,每一个结点有自己相应虫子数量,还有能量值,杀死虫子才能获得能量。一个士兵杀死20只虫子,问怎么样分配士兵才可以获得最多的能量,杀死父节点的虫子之后才能去扫荡子节点。 在这里,我对这道题目的理解是,树形dp,尤其是像这种类型的DP,其实就是背包问题的依赖背包变形。这道题目和这一道其实差不多的
背包九讲里讲过依赖背包,泛化物品。解决这种问题的方法就是从下而上,先解决子树的最优解,再由子树解决根的最优解。树形dp做到现在,基本上都是遵循这个规则,不同的就是状态转移方程发生一些变化
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;
int dp[105][105];//当前节点为i,小兵数为j得到的最大能量值
int n,m;
int bug[105];
int brain[105];
int head[105];
int tot;
int vis[105];
struct Node
{
int value;
int next;
}edge[105*2];
void add(int x,int y)
{
edge[tot].value=y;
edge[tot].next=head[x];
head[x]=tot++;
}
void dfs(int root)
{
vis[root]=1;
int term=(bug[root]+19)/20;
for(int i=term;i<=m;i++)
dp[root][i]=brain[root];
for(int i=head[root];i!=-1;i=edge[i].next)
{
int u=edge[i].value;
if(!vis[u])
{
dfs(u);
for(int j=m;j>=term;j--)
{
for(int k=1;j+k<=m;k++)
{
if(dp[u][k])
dp[root][j+k]=max(dp[root][j+k],dp[root][j]+dp[u][k]);
}
}
}
}
}
int main()
{
int x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==-1&&m==-1)
break;
memset(vis,0,sizeof(vis));
memset(head,-1,sizeof(head));
tot=0;
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&bug[i],&brain[i]);
}
for(int i=1;i<n;i++)
{
scanf("%d%d",&x,&y);
add(x,y);
add(y,x);
}
dfs(1);
if(m==0)
printf("0n");
else
printf("%dn",dp[1][m]);
}
}
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- PAT (Basic Level) Practice (中文)1022 D进制的A+B (20 分)
- [900]mysql字符串数字互转
- 技术分享 | Semi-join Materialization 子查询优化策略
- 前端性能和错误监控
- Network在单细胞转录组数据分析中的应用
- PAT (Basic Level) Practice (中文)1024 科学计数法 (20 分)
- PAT (Basic Level) Practice (中文)1053 住房空置率 (20 分)
- 解决vue+axios请求报错POST http: net::ERR_CONNECTION_REFUSED,在封装的请求中统一处理请求异常的问题
- PAT (Basic Level) Practice (中文)1025 反转链表 (25 分)
- Pytest+Allure接口自动化一些学习分享
- 数据结构___马踏棋盘详尽实现+报告+通俗易懂注释
- 使用elasticsearch-dump迁移elasticsearch集群数据
- PAT (Basic Level) Practice (中文)1027 打印沙漏 (20 分)
- PAT (Advanced Level) Practice 1002 A+B for Polynomials (25 分)
- 关于MySQL varchar类型最大值,原来一直都理解错了