C++ API 设计 03 序言

It is said that the people of Menggol lived in caves. A tribe‘s caves were connected to each other with paths. The paths were so designed that there was one and only one path to each cave. So the caves and the paths formed a tree. There was a main cave, which connected the world outside. The Menggolian always carved a map of the tribe‘s caves on the wall of the main cave.

Scientists have just discovered Menggolian‘s tribe. What a heart-stirring discovery! They are eager to explore it. Since the terrain is very complex and dangerous, they decide to send out a robot.

The robot will be landed into the main cave, where he will begin his adventure. It doesn‘t have to return to the main cave, because the messages of his exploration will be sent immediately to the scientists while he is on the way.

A robot can only walk x meters before it runs out of energy. So the problem arises: given the map of the tribe‘s caves and a set of x , how many caves can be explored at most?

Input 

There are multiple test cases in the input file. Each test case starts with a single number n (0n500) , which is the number of caves, followed by n - 1 lines describing the map. Each of the n - 1 lines contains three integers separated by blanks: i , j , and d (1d10000) . It means that the i -th cave‘s parent caveis the j -th cave and the distance is d meters. A parent cave of cave i is the first cave to enter on the path from i to the main cave. Caves are numbered from 0 to n - 1 . Then there is an integer q (1q1000) , which is the number of queries, followed by q lines. For one query, there is one integer x (0x5000000) , the maximum distance that the robot can travel. n = 0 indicates the end of input file.

Output 

For each test case, output q lines in the format as indicated in the sample output, each line contains one integer, the maximum number of caves the robot is able to visit.

Sample Input 

3 
1 0 5 
2 0 3 
3 
3 
10 
11 
0

Sample Output 

Case 1: 
2 
2 
3

题意：一个n结点的有根树，树边权均为正数，现在有q次询问，每次询问一个距离，问这个距离最多能经过多少点，同一结点多次经过算1次。

思路：树形dp，状态的表示为dp[u][j][0]表示根为u的树，遍历j个点，不返回u所需的最小距离，dp[u][j][1]则表示返回u的最小距离。这样的话就是根v的树查找k个点，根u的树查找j-k个点，之和就是j。状态转移方程为：

如此一来dp[u][j][1] = min{dp[v][k][1] + dp[u][j - k][1] + 2 * g[u][v].value);

dp[u][j][0] = min(dp[v][k][0] + dp[u][j - k][1] + g[u][v].value, dp[v][k][1] + dp[u][j - k][0] + 2 * g[u][v].value);

代码：

#include <stdio.h>
#include <string.h>
#include <vector>
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N = 505;
int n, q, u, v, value, num, i, vis[N], dp[N][N][2], sum[N];
struct Node {
	int v, value;
	Node() {}
	Node(int vv, int vva) {v = vv; value = vva;}
};
vector<Node> g[N];

void dfs(int u) {
	vis[u] = sum[u] = 1;
	dp[u][1][0] = dp[u][1][1] = 0;
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i].v, value = g[u][i].value;
		if (vis[v]) continue;
		dfs(v);
		sum[u] += sum[v];
		for (int j = sum[u]; j > 1; j--) {
			for (int k = 0; k <= j && k <= sum[v]; k++) {
				dp[u][j][1] = min(dp[u][j][1], dp[u][j - k][1] + dp[v][k][1] + 2 * value);
				dp[u][j][0] = min(dp[u][j][0], dp[u][j - k][1] + dp[v][k][0] + value);
				dp[u][j][0] = min(dp[u][j][0], dp[u][j - k][0] + dp[v][k][1] + 2 * value);
			}
		}
	}
}

int main() {
	int cas = 0;
	while (~scanf("%d", &n) && n) {
		memset(vis, 0, sizeof(vis));
		memset(g, 0, sizeof(g));
		for (i = 0; i < n - 1; i++) {
			scanf("%d%d%d", &v, &u, &value);
			g[u].push_back(Node(v, value));
			g[v].push_back(Node(u, value));
		}
		memset(dp, INF, sizeof(dp));
		dfs(0);
		scanf("%d", &q);
		printf("Case %d:\n", ++cas);
		while (q--) {
			scanf("%d", &num);
			for (int i = n; i >= 1; i--) {
				if (dp[0][i][0] <= num || dp[0][i][1] <= num) {
					printf("%d\n", i);
					break;
				}
			}
		}
	}
	return 0;
}