1053 Path of Equal Weight (30分)
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 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
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",¤t);
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);
}
