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)
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! "); 86 continue; 87 } 88 t -= Mint; 89 dfs2(1, -1); 90 printf("%d ", dp[1][t]); 91 } 92 return 0; 93 }