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 Lis 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.



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, 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

回溯+剪枝

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 105;
int n,m,s;
vector<int> T[maxn];
vector<int> res;
int weight[maxn];
bool cmp(int x,int y){return weight[x] > weight[y];}
void dfs(int x,int w){
if(w == s){
if(T[x].size() == 0){
for(int i=0;i<res.size();i++){
if(i != 0)cout<<" ";
cout<<res[i];
}cout<<endl;
}
return;
}
sort(T[x].begin(),T[x].end(),cmp);
for(int i=0;i<T[x].size();i++){
int v = T[x][i],nw = w+weight[v];
if(nw <= s){
res.push_back(weight[v]);
dfs(v,nw);
res.pop_back();
}
}
}
int main(){
cin>>n>>m>>s;
for(int i=0;i<n;i++)cin>>weight[i];
for(int i=0;i<m;i++){
int id,k,t;
cin>>id>>k;
for(int j=0;j<k;j++){
cin>>t;
T[id].push_back(t);
}
}
res.push_back(weight[0]);
dfs(0,weight[0]);
return 0;
}