HDOJ 4276 The Ghost Blows Light(树形DP)

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

 

思路

1. 经典树形 DP 题目

2. 动规数组定义: dp[i][j] 表示从节点 i 出发使用最大花费为 j 后重新回到节点 i 能够得到的最大价值

3. 状态转移方程: dp[i][j] = dp[c1][k1] + dp[c2][k2] + ... + dp[cn][kn], c1, c2 ... cn 为节点 i 的孩子, k1, k2 ... kn 是在其对应的孩子节点分配的时间,

　　k1+k2+...kn <= j

4. 盗贼必须要跑出去, 因此最短路径上的时间需要预支. 以最短路径上的点(u)为树根求出以 u 为根的树分配 j 时间能够得到的最大收益 dp[u][j]

5. 然后再计算, 如何在最短路径上的节点分配时间使总收益最大, 这也是一步 DP, 具体来说, 这依然是一步树形 DP

6. 实现分为 3 部分, 第一部分使用 BFS(bellmanford, dijkstra) 算出最短路径的耗费以及最短路径上的所有点; 第二部分是 (4); 第三部分是 (5)

 

代码 from kedebug

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
#define max(a,b)    (((a) > (b)) ? (a) : (b))
const int INF = 1e9;

struct Node {
    int y, w;
    Node(int _y, int _w) : y(_y), w(_w) { }
};

vector<Node> map[110];
int dp[110][510];   // dp[i][j] 从i出发又返回i，最大花费为j时所取得的价值
int val[110], mark[110];

int bfs(int s, int e)
{
    int dist[110], f[110];
    for (int i = 0; i < 110; ++i)
        dist[i] = INF;
    f[s] = -1;
    dist[s] = 0;
    queue<int> q;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i = 0; i < map[u].size(); ++i)
        {
            Node &v = map[u][i];
            if (dist[v.y] > dist[u] + v.w)
            {
                f[v.y] = u;
                dist[v.y] = dist[u] + v.w;
                q.push(v.y);
            }
        }
    }
    for (int i = e; i != -1; i = f[i])
        mark[i] = 1;
    return dist[e];
}

void dfs(int u, int pre, int mval)
{
    dp[u][0] = val[u];
    for (int i = 0; i < map[u].size(); ++i)
    {
        int y = map[u][i].y;
        int w = map[u][i].w;
        if (y == pre || mark[y])
            continue;
        dfs(y, u, mval);

        for (int j = mval; j >= 0; --j)
            for (int k = 0; k <= j-2*w; ++k)
                if (dp[u][j-k-2*w] != -1 && dp[y][k] != -1)
                    dp[u][j] = max(dp[u][j], dp[y][k] + dp[u][j-k-2*w]);
    }
}

int main()
{
    int n, t;
    while (scanf("%d %d", &n, &t) == 2)
    {
        memset(map, 0, sizeof(map));
        memset(dp, -1, sizeof(dp));
        memset(mark, 0, sizeof(mark));

        int a, b, c;
        for (int i = 1; i < n; ++i)
        {
            scanf("%d %d %d", &a, &b, &c);
            map[a].push_back(Node(b, c));
            map[b].push_back(Node(a, c));
        }
        for (int i = 1; i <= n; ++i)
            scanf("%d", &val[i]);

        int tt = bfs(1, n);
        if (tt > t)
        {
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
            continue;
        }
        for (int i = 1; i <= n; ++i)
            if (mark[i])
                dfs(i, -1, t - tt);

        int dp2[510], tmax = t - tt;
        memset(dp2, -1, sizeof(dp2));
        dp2[0] = 0;
        for (int i = 1; i <= n; ++i)
        {
            if (!mark[i])
                continue;
            for (int j = tmax; j >= 0; --j)
            {
                for (int k = 0; k <= j; ++k)
                    if (dp2[j-k] != -1 && dp[i][k] != -1)
                        dp2[j] = max(dp2[j], dp2[j-k] + dp[i][k]);
                    
            }
        }
        int ans = 0;
        for (int i = 0; i <= tmax; ++i)
            if (ans < dp2[i])
                ans = dp2[i];
        printf("%d\n", ans);
    }
    return 0;
}