HDOJ 4276 鬼吹灯 (树形DP)

 

总结

1. dp[u][k] 是恰好用完 k 时间的最大价值

2. dfs(u, pre, vmax) 是计算 dp[u][1...vmax] 的值

3. dfs 内部遍历, j 从大到小, 防止重复计算 f(j) = g(j-k)

而 k 是从小到大, 因为在 134 行,  dp[j] = max(dp[j], dp[j-k]+dp'[k]) 比较特殊, k 可以等于 0. 导致 dp[j] 被重复计算

 

代码

/*
 * source.cpp
 *
 *  Created on: Apr 6, 2014
 *      Author: sangs
 */


#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <algorithm>
#include <math.h>
using namespace std;

class Road {
public:
	int to, cost;
	Road(int _to, int _cost):to(_to), cost(_cost) {}
	Road() {
		to = cost = 0;
	}
	bool operator <(const Road &rhs) const {
		return this->cost < rhs.cost;
	}
};

vector<Road> graph[1000];
int val[1000];
int dp[1000][1000];

const int INF = 0X3F3F3F3F;
int dist[1000];
int father[1000];
bool mark[1000]; // mark those nodes must pass
bool visited[1000];
/*
 * calculate the shortest path from 1 to n
 * as well as those nodes in shortest path
 * SPFA algorithm
 */
int bfs(int n) {
	queue<int> record;

	memset(visited, 0, sizeof(visited));
	memset(mark, 0,    sizeof(mark));
	memset(dist, 0x3f, sizeof(dist));

	record.push(1);
	dist[1] = 0;
	father[1] = -1;

	while(!record.empty()) {
		int thisNode = record.front();
		record.pop();

		for(size_t i = 0; i < graph[thisNode].size(); i ++) {
			int j = graph[thisNode][i].to;
			//cout << j << " " << dist[i] << " " << graph[thisNode][i].cost << endl;

			if(dist[thisNode] + graph[thisNode][i].cost < dist[j]) {
				dist[j] = dist[thisNode] + graph[thisNode][i].cost;
				father[j] = thisNode;
				record.push(j);
			}
		}
	}

	for(int i = n; i != -1; i = father[i]) {
		mark[i] = 1;
	}
	return dist[n];
}
/*
 * tree_dp to calculate dp[u][1...vmax]
 * dp[u][k] is the maximum income when assigned k time to node u and get back to u
 * pre is u's father
 */
void tree_dp(int u, int pre, int vmax) {
	dp[u][0] = val[u];

	for(size_t i = 0; i < graph[u].size(); i ++) { // enum v's child

		int index = graph[u][i].to;
		if(index == pre) continue;
		if(mark[index]) continue;

		tree_dp(index, u, vmax);

		int cost = graph[u][i].cost;
		for(int j = vmax; j >= 0; j --) {
			for(int k = 0; k <= j-2*cost; k ++) {
				dp[u][j] = max(dp[u][j], dp[u][j-k-2*cost]+dp[index][k]);
			}
		}
	}
}

int main() {
	freopen("input.txt", "r", stdin);
	int n, T;
	while(scanf("%d%d", &n, &T) != EOF) {
		for(int i = 0; i <= n+10; i ++)
			graph[i].clear();

		for(int i = 0; i < n-1; i ++) {
			int from, to, cost;
			scanf("%d%d%d", &from, &to, &cost);
			graph[from].push_back(Road(to, cost));
			graph[to].push_back(Road(from, cost));
		}

		for(int i = 1; i <= n; i ++)
			scanf("%d", val+i);


		int cutTime = bfs(n);
		T -= cutTime;

		memset(dp, 0, sizeof(dp));
		dp[n+1][0] = 0;
		for(int i = 1; i <= n; i ++) {
			if(mark[i] == 0) continue;

			tree_dp(i, n+1, T);

			for(int j = T; j >= 0; j --) {
				for(int k = 0; k <= j; k ++) {
					//for(int k = j; k >= 0; k --) {
					dp[n+1][j] = max(dp[n+1][j], dp[n+1][j-k] + dp[i][k]);
				}
			}
		}
		int res = 0;
		for(int i = 0; i <= T; i ++)
			res = max(res, dp[n+1][i]);
		printf("%d\n", res);
	}

    return 0;
}