(PAT 1053) Path of Equal Weight (树的深度优先遍历+剪枝)

喜欢ヅ旅行 2022-03-21 14:42 144阅读 0赞

1053 Path of Equal Weight （30 point(s)）

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The **weight of a path from R to L**is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in the following figure: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: \{10 5 2 7\}, \{10 4 10\}, \{10 3 3 6 2\} and \{10 3 3 6 2\}, which correspond to the red edges in the figure.

![212][]

### Input Specification: ###

Each input file contains one test case. Each case starts with a line containing 0<N≤100, the number of nodes in a tree, M(<N), the number of non-leaf nodes, and 0<S<230, the given weight number. The next line contains N positive numbers where Wi (<1000) corresponds to the tree node Ti. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where `ID` is a two-digit number representing a given non-leaf node, `K` is the number of its children, followed by a sequence of two-digit `ID`'s of its children. For the sake of simplicity, let us fix the root ID to be `00`.

### Output Specification: ###

For each test case, print all the paths with weight S in **non-increasing** order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence \{A1,A2,⋯,An\} is said to be **greater than** sequence \{B1,B2,⋯,Bm\} if there exists 1≤k<min\{n,m\} such that Ai=Bi for i=1,⋯,k, and Ak+1>Bk+1.

### Sample Input: ###

20 9 24
    10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
    00 4 01 02 03 04
    02 1 05
    04 2 06 07
    03 3 11 12 13
    06 1 09
    07 2 08 10
    16 1 15
    13 3 14 16 17
    17 2 18 19

### Sample Output: ###

10 5 2 7
    10 4 10
    10 3 3 6 2
    10 3 3 6 2

解题思路:

深度优先遍历树,通过vector保存路径

遍历到一个结点,如果该结点不是叶节点且加上该节点的权重后大于等于指定值,就剪去该段路径

这里的难点是最后的排序问题,我们只需把每个结点的叶节点从左到右按照从大到小的顺序排序,由于我们遍历是先序遍历的,所以我们遍历的路径结点大小是从大到小的

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    using namespace std;
    int N, M;
    long long S;
    struct WNode
    {
    	int wegiht;
    	vector<int> childs;
    }WTree[110];
    
    vector<int> tempPath;
    vector<int> Pathes[10001];
    
    bool cmp(int a,int b) {
    	return WTree[a].wegiht > WTree[b].wegiht;
    }
    
    int index = 0;
    void DFS_W(int root,long long ans) {
    	ans += WTree[root].wegiht;
    	if (WTree[root].childs.empty()) {
    		if (ans == S) {
    			tempPath.push_back(WTree[root].wegiht);
    			Pathes[index++] = tempPath;
    			tempPath.pop_back();
    		}
    		return;
    	}
    	else {
    		if (ans >= S) return;
    		else {
    			tempPath.push_back(WTree[root].wegiht);
    			for (int i = 0; i < WTree[root].childs.size(); ++i) {
    				DFS_W(WTree[root].childs[i], ans);
    			}
    			tempPath.pop_back();
    		}
    	}
    }
    
    int main() {
    	scanf("%d %d %lld", &N, &M, &S);
    	for (int i = 0; i < N; ++i) {
    		scanf("%d", &WTree[i].wegiht);
    	}
    	for (int i = 0; i < M; ++i) {
    		int root, sums, node;
    		scanf("%d %d", &root, &sums);
    		for (int j = 0; j < sums; ++j) {
    			scanf("%d", &node);
    			WTree[root].childs.push_back(node);
    		}
    		sort(WTree[root].childs.begin(), WTree[root].childs.end(), cmp);  //确保从左到右是从大到小排序
    	}
    	DFS_W(0,0);
    
    	for (int i = 0; i < index; ++i) {
    		for (int j = 0; j < Pathes[i].size(); ++j) {
    			printf("%d", Pathes[i][j]);
    			if (j < Pathes[i].size() - 1) printf(" ");
    		}
    		printf("\n");
    	}
    
    	system("PAUSE");
    	return 0;
    }

[212]: /images/20220321/7548411cfee64e70b46a1d61c1f40552.png