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# PAT (Advanced Level) Practice 1053 Path of Equal Weight (30 分) 凌宸1642

2周前 (04-08) 4次浏览

## PAT (Advanc++ed Level) Practice 1053 Path of Equal Weight (30 分) 凌宸1642

### 题目描述：

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.

### Input Spec++ification (输入说明):

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`.

``````ID K ID[1] ID[2]…ID [K]
``````

### Output Spec++ification (输出说明):

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 (样例输入):

``````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 (样例输出):

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

### The Codes:

``````#include<bits/stdc++.h>
using namespace std ;
#define MAX 110
struct node{
int weight ; // 权值
vector<int> child ; // 指针域，所有孩子结点
}Node[MAX] ;
bool cmp(int a , int b){
return Node[a].weight > Node[b].weight ; // 将子结点按照 权值降序排列
}
int n , m , s ;
int path[MAX] ;
void dfs(int index , int nowNode , int sumW){
if(sumW > s) return ; //如果权值大于 s ，直接返回
if(sumW == s){
if(Node[index].child.size() != 0) return ; // 有子结点，说明未到达叶子结点， 直接返回
for(int i = 0 ; i < nowNode ; i ++){ // 此时 path 中已有一条完整的路径，输出
printf("%d%c" , Node[path[i]].weight , (i == nowNode - 1)?'n':' ') ;
}
return ;
}
for(int i = 0 ; i < Node[index].child.size() ; i ++){ // 对于当前结点的所有子结点
int ch = Node[index].child[i] ; // 获得当前结点的第 i 个子结点
path[nowNode] = ch ; // 将第 i 个子结点加入 path 中
dfs(ch , nowNode + 1 , sumW + Node[ch].weight) ; // 对变化后，再一次深搜
}
}
int main(){
scanf("%d%d%d" , &n , &m , &s) ; // 输入 n ，m ，s
for(int i = 0 ; i < n ; i ++){
scanf("%d" , &Node[i].weight) ; // 输入每个结点的权值
}
int k , l  , t ;
for(int i = 0 ; i < m ; i ++){ // m 行输入
scanf("%d%d" , &k , &l) ; // 结点 k 以及其子结点个数 l
for(int j = 0 ; j < l ; j ++){
scanf("%d" , &t) ;
Node[k].child.push_back(t) ; // 存储结点 k 的每一个子结点
}
sort(Node[k].child.begin() , Node[k].child.end() , cmp ) ; // 将子结点按权值降序排列
}
path[0] = 0 ; // 起点为 根结点
dfs(0 , 1 , Node[0].weight ) ; // dfs初始
return 0;
}

``````