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HDU 1011 Starship Troopers（樹形dp+背包）

編輯：關於C++

Starship Troopers

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13109 Accepted Submission(s): 3562

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

Sample Output

50
7

Author XU, Chuan
Source ZJCPC2004

/*

題意:起點1，然後n個點，m個人，每一個節點有 cost 和va,一個人最多消滅20個cost
     但是不能走一個節點兩遍，消滅一個節點的全部cost後得到這個節點的va,
     而且消滅他的人不能再用了，求這m個人最多可以得到多少 va

思路: 樹形dp+背包，背包部分在代碼寫出


*/

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define INF 0x3f3f3f3f
#define N 155

vectorg[N];

int dp[N][N];
int cost[N],va[N];
int n,m;
int vis[N];

void dfs(int u)
{
    int i,j;
    int lim=(cost[u]+19)/20;
    for(i=lim;i<=m;i++)
	   dp[u][i]=va[u];

	vis[u]=1;
	for(i=0;i=lim;v--)     //假設這個點的人數最大為m,
	                                 // 那麼依次求人數在m~lim時的最大情況(背包)
			for(int co=1;v-co>=lim;co++)   //枚舉分幾個人給to節點
			  dp[u][v]=max(dp[u][v],dp[u][v-co]+dp[to][co]);
	}
}

int main()
{
    int i,j;
    while(scanf(%d%d,&n,&m))
	{
		if(n==-1&&m==-1) break;
		for(i=1;i<=n;i++)
			scanf(%d%d,&cost[i],&va[i]);
		for(i=0;i<=n;i++)
			g[i].clear();

		int u,v;
		for(i=1;i