The Ghost Blows Light
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1567 Accepted Submission(s): 484
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
Source
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liuyiding
题意:给你两个数n, T。 表示有n个房间, 和时间T.
然后有n - 1条无向边, a, b, c 表示 a 与 b 的距离是 c
然后有n个数字, 表示每个房间的宝藏数量。
问从房间1出发, 到房间n出去。 问能拿到多少宝藏。 如果出不出就输出 “……”;
思路: 树形DP
由于是一颗树, 每两个节点只有唯一路径, 所以要先把1~n的路径先搜索出来, 然后把边都赋值为0, 宝藏都赋值为0。
然后再树形DP。
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <vector> using namespace std; //树形DP const int V = 100 + 50; const int MaxN = 500 + 5; const int mod = 1000000000 + 7; int n, T, pri[V], Map[V][V], sum_path; int dp[V][MaxN];// dp[i][j] 表示 第i个房间花费j时间能拿到的最大的宝藏 bool visit[V]; void dfs(int index) { visit[index] = true; for(int i = 1; i <= n; ++i) if(!visit[i] && Map[index][i] >= 0) { dfs(i); for(int j = T; j >= Map[index][i] * 2; --j) for(int k = 0; k <= j - Map[index][i] * 2; ++k) dp[index][j] = max(dp[index][j], dp[index][j - Map[index][i] * 2 - k] + dp[i][k] + pri[i]); } } bool dfs_path(int index, int len, int p) { visit[index] = true; if(index == n) { pri[index] = 0; sum_path = p; T -= len; return true; } for(int i = 1; i <= n; ++i) if(!visit[i] && Map[index][i] > 0) { if(dfs_path(i, len + Map[index][i], p + pri[i])) { Map[index][i] = Map[i][index] = 0; pri[index] = 0; return true; } } return false; } int main() { int i, j; while(~scanf("%d%d", &n, &T)) { memset(Map, -1, sizeof(Map)); memset(dp, 0, sizeof(dp)); memset(visit, false, sizeof(visit)); sum_path = 0; for(i = 0; i < n - 1; ++i) { int a, b, c; scanf("%d%d%d", &a, &b, &c); Map[a][b] = c; Map[b][a] = c; } for(i = 1; i <= n; ++i) scanf("%d", &pri[i]); dfs_path(1, 0, pri[1]); if(T < 0) printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n"); else { memset(visit, false, sizeof(visit)); dfs(1); printf("%d\n", dp[1][T] + sum_path); } } }