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 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 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<2​30​​, 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 {A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B​1​​,B​2​​,⋯,B​m​​} if there exists 1≤k<min{n,m} such that A​i​​=B​i​​ for 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

Special thanks to Zhang Yuan and Yang Han for their contribution to the judge's data.

 

 

#include "iostream"

#include "vector"

#include "algorithm"

using namespace std;

const int maxn=100;

int path[maxn];

struct node{

int data;

int weight;

vector<int> child;

}Node[maxn];

 

bool comp(int x,int y){

return Node[x].weight>Node[y].weight;

}

 

void dfs(int node,int layer,int weight){

path[layer]=node;

weight=weight-Node[node].weight;

if (weight<0) {

return;

}

if (weight==0) {

if (Node[node].child.size()!=0) {

return;

}

else{

for (int i=0; i<=layer; i++) {

if (i!=layer) {

printf("%d ",Node[path[i]].weight);

}

else{

printf("%d",Node[path[i]].weight);

}

}

printf("\n");

}

}

for (int i=0; i<Node[node].child.size(); i++) {

dfs(Node[node].child[i], layer+1, weight);

}

}

 

int main(){

int sum,Noleaf,aim;

scanf("%d %d %d",&sum,&Noleaf,&aim);

for (int i=0; i<sum; i++) {

int weight;

scanf("%d",&weight);

Node[i].weight=weight;

}

for (int i=0; i<Noleaf; i++) {

int current;

int number;

scanf("%d",&current);

scanf("%d",&number);

for (int j=0; j<number; j++) {

int s;

scanf("%d",&s);

Node[current].child.push_back(s);

}

sort(Node[current].child.begin(), Node[current].child.end(),comp);

}

dfs(0, 0, aim);

}