【PAT】1053. Path of Equal Weight (30)

原創

2020-02-21 19:30

Given a non-empty tree with root R, and with weight W_i assigned to each tree node T_i. 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 Figure 1: 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 Figure 1.

Figure 1

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 W_i (<1000) corresponds to the tree node T_i. 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, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, and A_k+1 > B_k+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

分析：找出從根到葉子節點，花銷爲給定值的路徑，並按照一定次序排列。

正確代碼如下：

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

struct node{
	int id;
	int wei;
	int pre;
};
vector<node> tree;
vector<vector<node> > path;
int wei;
void calWei(vector<node>& v){
	int i;
	for(i=0; i<v.size(); i++){
		vector<node> possiblePath;
	 	node tmp;
		tmp = v[i];
		int w = 0;
	 	while(tmp.id!= 0){
	 		w += tmp.wei;
	 		possiblePath.push_back(tmp);
	 		tmp = tree[tmp.pre];
		}
		w += tree[0].wei;
		if(w == wei){
			possiblePath.push_back(tree[0]);
			path.push_back(possiblePath);
		}		
	}
}

bool cmp(vector<node> a, vector<node> b){
	int i = a.size()-1;
	int j = b.size()-1;
	for(; i>=0&&j>=0; i--,j--){
		if(a[i].wei != b[j].wei){
			return a[i].wei>b[j].wei;
		}	
	}	
	if(i==0 && j!=0){
		return true;
	}else if(i!=0 && j==0){
		return false;
	}
	return false;
}

int main(int argc, char** argv) {
	int n, m, i;
	scanf("%d%d%d",&n,&m,&wei);
	tree.resize(n);
	for(i=0; i<n; i++){
		tree[i].id= i;
		scanf("%d",&tree[i].wei);
	} 
	set<int> nonLeaf;
	for(i=0; i<m; i++){
		int index, num, t, j;
		scanf("%d%d",&index,&num);
		nonLeaf.insert(index);
		for(j=0; j<num; j++){
			scanf("%d",&t);
			tree[t].pre = index;
		}
	}
	vector<node> leafNode;
	for(i=0; i<n; i++){
		if(nonLeaf.count(i)==0){
			leafNode.push_back(tree[i]);
		}
	}
	calWei(leafNode);
	sort(path.begin(), path.end(), cmp);
	for(i=0; i<path.size(); i++){
		for(int j=path[i].size()-1; j>=0; j--){
			if(j==0){
				printf("%d\n",path[i][j].wei);
			}else{
				printf("%d ",path[i][j].wei);
			}
		}		
	}
	return 0;
}

解法二：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> wei;

struct node{
	int pre, wei;
};

vector<node> tree;
int countWei(int leaf, vector<int> &vec){
	int addr=leaf, total=0;
	while(true){
		vec.push_back(wei[addr]);
		total += wei[addr];
		if(addr == 0)
			break;
		addr = tree[addr].pre;
	}
	return total;
}

int main(int argc, char** argv) {
	int N, M, S;
	scanf("%d%d%d",&N,&M,&S);
	int i;
	wei.resize(N);
	for(i=0; i<N; i++){
		scanf("%d",&wei[i]);
	}
	int fa, num, t;
	tree.resize(N);
	vector<int> isLeaf(N,1);
	for(i=0; i<M; i++){
		scanf("%d%d",&fa,&num);
		isLeaf[fa] = 0;
		for(int j=0; j<num; j++){
			scanf("%d",&t);
			tree[t].pre = fa;			
		}
	}
	vector<vector<int> >result;
	for(i=0; i<N; i++){
		if(isLeaf[i]==1){
			vector<int> tmp;
			int wei = countWei(i, tmp);
			if(wei == S){
				reverse(tmp.begin(), tmp.end());
				result.push_back(tmp);
			}
		}
	}
	sort(result.begin(), result.end());
	for(i=result.size()-1; i>=0; i--){
		for(int j=0; j<result[i].size(); j++){
			if(j==0)
				printf("%d",result[i][j]);
			else
				printf(" %d",result[i][j]);
		}
		printf("\n");
	}
	return 0;
}