PAT甲级1053. Path of Equal Weight (30)

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 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.



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=Bifor i=1, … k, andAk+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

#include <cstdio>
using namespace std;
#include <vector>
#include <algorithm>

int N,M,S;
vector<int> W; //weight of nodes
vector<vector<int> > G; //邻接表
int vis[110];
vector<int> path;
vector<vector<int> > ans;
int currentSum=0;

bool compare(vector<int> &a,vector<int> &b){
for(int i=0;i<a.size()&&i<b.size();i++){
if(a[i]!=b[i]) return a[i]>b[i];
}
return a.size()>b.size();
}

void DFS(int s){
vis[s]=1;
if(G[s].size()==0&¤tSum==S){
ans.push_back(path);
return;
}
if(G[s].size()>0&¤tSum>S) return;
for(int i=0;i<G[s].size();i++){
currentSum+=W[G[s][i]];
path.push_back(W[G[s][i]]);

DFS(G[s][i]);

currentSum-=W[G[s][i]];
path.pop_back();

}
}

int main(){
scanf("%d %d %d",&N,&M,&S);

W.resize(N);
for(int i=0;i<N;i++) scanf("%d",&W[i]);

G.resize(N);
int ID,K,child;
for(int i=0;i<M;i++){
scanf("%d %d",&ID,&K);
for(int j=0;j<K;j++) {
scanf("%d",&child);
G[ID].push_back(child);
}
}

path.push_back(W[0]);
currentSum=W[0];
DFS(0);

sort(ans.begin(),ans.end(),compare);
for(int i=0;i<ans.size();i++){
for(int j=0;j<ans[i].size();j++){
if(j<ans[i].size()-1) printf("%d ",ans[i][j]);
else printf("%d\n",ans[i][j]);
}
}

return 0;
}