PAT 1053 Path of Equal Weight (30 分)
1053 Path of Equal Weight (30 分)
Given a non-empty tree with root R, and with weight
assigned to each tree node
. 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.
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<
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 {
,
,⋯,
} is said to be greater than sequence {
,
,⋯,
} if there exists 1≤k<min{n,m} such that
=
for i=1,⋯,k, and
>$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
解析
問一條路徑的和是否為Sum。好像LeetCode也有一道很類似的。112:Path Sum
這題我用的是DFS。但是常規的DFS是不能滿足我的需要的。所以我寫了一個DFShelp函式。
void DFShelp(vector<int>& Path,int len,int n, int s) {
if (s == Node[n].weight && Node[n].child.empty()) {
for (int i = 0; i < len; i++)
printf("%d ", Node[Path[i]].weight);
printf("%d\n",Node[n].weight);
}
else if (s < Node[n].weight) //回溯剪枝
return;
else { //繼續深入
Path[len] = n;
for (auto x : Node[n].child) {
DFShelp(Path, len + 1, x, s - Node[n].weight);
}
}
}
因為輸出的路徑權重 要按降序,所以可以在DFS之前:給每一個結點的子節點排序。
Code:
#include<algorithm>
#include<functional>
#include<cstdio>
#include<iostream>
#include<queue>
#include<string>
#include<cmath>
const int maxn = 101;
using namespace std;
int LayerTable[101]{0};
struct node {
int weight;
vector<int> child;
node() :weight(0) { }
}Node[maxn];
void DFShelp(vector<int>& Path,int len,int n, int s) {
if (s == Node[n].weight && Node[n].child.empty()) {
for (int i = 0; i < len; i++)
printf("%d ", Node[Path[i]].weight);
printf("%d\n",Node[n].weight);
}
else if (s < Node[n].weight)
return;
else {
Path[len] = n;
for (auto x : Node[n].child) {
DFShelp(Path, len + 1, x, s - Node[n].weight);
}
}
}
void DFS(int n,int s) {
vector<int> Path(101,0);
DFShelp(Path,0,n,s);
}
int main()
{
int N, M, S;
scanf("%d %d %d", &N, &M, &S);
for (int i = 0; i < N; i++)
scanf("%d", &Node[i].weight);
for (int i = 0; i < M; i++) {
int father, K, son;
scanf("%d %d", &father, &K);
for (int i = 0; i < K; i++) {
scanf("%d", &son);
Node[father].child.push_back(son);
}
}
for (int i = 0; i < N; i++)
sort(Node[i].child.begin(), Node[i].child.end(), [](int a, int b) {
return Node[a].weight > Node[b].weight; });
DFS(0,S);
}