HDU 4276 - The Ghost Blows Light

Posted 2020-08-04 stort

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3345    Accepted Submission(s): 1040

Problem Description

　　My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road should cost some time). There is exactly one route between any two rooms, and each room contains some treasures. Now I am located at the 1st room and the exit is located at the Nth room.
Suddenly, alert occurred! The tomb will topple down in T minutes, and I should reach exit room in T minutes. Human beings die in pursuit of wealth, and birds die in pursuit of food! Although it is life-threatening time, I also want to get treasure out as much as possible. Now I wonder the maximum number of treasures I can take out in T minutes.

 

Input

 

　　There are multiple test cases.
The first line contains two integer N and T. (1 <= n <= 100, 0 <= T <= 500)
Each of the next N - 1 lines contains three integers a, b, and t indicating there is a road between a and b which costs t minutes. (1<=a<=n, 1<=b<=n, a!=b, 0 <= t <= 100)
The last line contains N integers, which Ai indicating the number of treasure in the ith room. (0 <= Ai <= 100)

 

Output

 

　　For each test case, output an integer indicating the maximum number of treasures I can take out in T minutes; if I cannot get out of the tomb, please output "Human beings die in pursuit of wealth, and birds die in pursuit of food!".

 

Sample Input

 

5 10

1 2 2

2 3 2

2 5 3

3 4 3

1 2 3 4 5

 

Sample Output

 

11

 

题意：

　　有一棵树，有n个节点，每个节点有一个权值，每一条边有一个时间，在经过的时间不超过给定时间m的情况下，求从1点到n点的路径的最大权值

思路：

　　对于一棵树只有一条 从根节点到某节点的最短路，所以先判断这条最短路是不是符合条件，符合的话，把这条路上的点记为0，然后就是树形DP求解，

因为要走其他点，必须再回来，所以回来时要乘以二所经过的路径

这是个背包问题：

状态转移方程为，dp[i][j]=max(dp[i][j],dp[i][k]+dp[i][j-2*val-k])

AC代码：

 1 # include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int MAX = 120;
 5 struct node
 6 {
 7     int to;
 8     int next;
 9     int t;
10 }tree[MAX];
11 int head[MAX];
12 int tol = 0;
13 int w[MAX];
14 int n, t;
15 int Mint;
16 int dp[MAX][600];
17 
18 void add(int a, int b, int val)
19 {
20     tree[tol].to = b;
21     tree[tol].next = head[a];
22     tree[tol].t = val;
23     head[a] = tol++;
24 }
25 
26 bool dfs1(int root, int f)
27 {
28     if(root == n)
29         return true;//找到了
30     for(int i = head[root]; i != -1; i = tree[i].next)
31     {
32         int son = tree[i].to;
33         if(son == f)
34             continue;
35         if(dfs1(son, root))
36         {
37             Mint += tree[i].t;
38             tree[i].t = 0;
39             return true;
40         }
41     }
42     return false;
43 }
44 
45 void dfs2(int root, int f)
46 {
47     for(int i = 0; i <= t; i++)
48         dp[root][i] = w[root];
49     for(int i = head[root]; i != -1; i = tree[i].next)
50     {
51         int son = tree[i].to;
52         if(son == f)
53             continue;
54         dfs2(son, root);
55         int cost = tree[i].t * 2;
56         for(int j = t; j >= cost; j--)
57         {
58             for(int k = 0; k <= j - cost; k++)
59                 dp[root][j] = max(dp[root][j], dp[root][j - k - cost] + dp[son][k]);
60         }
61     }
62 }
63 int main()
64 {
65     while(scanf("%d%d", &n, &t) != EOF)
66     {
67         memset(head, -1, sizeof(head));
68         memset(dp, 0, sizeof(dp));
69         tol = 0;
70         Mint = 0;
71         int a, b, val;
72         for(int i = 1; i < n; i++)
73         {
74             scanf("%d%d%d", &a, &b, &val);
75             add(a, b, val);
76             add(b, a, val);
77         }
78         for(int i = 1; i <= n; i++)
79         {
80             scanf("%d", &w[i]);
81         }
82         dfs1(1, -1);// 看看直接走去的最短时间是多少，
83         if(Mint > t)
84         {
85             printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\\n");
86             continue;
87         }
88         t -= Mint;
89         dfs2(1, -1);
90         printf("%d\\n", dp[1][t]);
91     }
92     return 0;
93 }

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