Starship Troopers
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4118 Accepted Submission(s): 1076
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.
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.
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
Author
XU, Chuan
Source
樹形DP
定義dp[i][j]默示根結點為i時，用掉j個士兵獲得的最大possible。
dp[i][j] = max（dp[i][j]， dp[i][j-k] + dp[son[i]][k]）;
遞歸的求出dp[son[i]][k]的值。最后成果就是dp[1][m];
題目大意是有n個房間組成一棵樹，你有m個士兵，從1號房間開始讓士兵向相鄰的房間出發，每個房間里有一個代價，代價是值/20個士兵，
同時有一個價值，問你花費這m個士兵可以得到的最大價值是多少，
#include<stdio.h>
#include<iostream>
using namespace std;
const int MAXN=110;
int N,M;
struct Node
{
int number,p;
};
Node node[MAXN];//記錄結點
int dp[MAXN][MAXN];//DP，dp[i][j]表示跟結點為i時，用掉j個士兵獲得的最大值
int adj[MAXN][MAXN];//存樹
bool vis[MAXN];//訪問標記
void dfs(int root)//DFS
{
vis[root]=true;//已經訪問
int num=(node[root].number+19)/20;//獲得該結點需要的士兵數目
for(int i=num;i<=M;i++) dp[root][i]=node[root].p;
for(int i=1;i<=adj[root][0];i++)
{
int u=adj[root][i];
if(vis[u]) continue;
dfs(u);
for(int j=M;j>=num;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 b,e;
while(scanf("%d%d",&N,&M))
{
if(N==-1&&M==-1) break;
memset(vis,false,sizeof(vis));
memset(dp,0,sizeof(dp));
memset(adj,0,sizeof(adj));
for(int i=1;i<=N;i++)
scanf("%d%d",&node[i].number,&node[i].p);
for(int i=1;i<N;i++)//存圖
{
scanf("%d%d",&b,&e);
adj[b][0]++;
adj[b][adj[b][0]]=e;
adj[e][0]++;
adj[e][adj[e][0]]=b;
}
if(M==0)//這個必需要，有代價為0的房間，M=0則無法獲得
printf("0\n");
else
{
dfs(1);
printf("%d\n",dp[1][M]);
}
}
return 0;
}
HDU 第235題了，紀念下。
離500題的目標越來越近了~~~~~~