1053. Path of Equal Weight (30) PAT甲级刷题

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.


Figure 1Input 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
思路：较简单的dfs回溯剪枝，剪枝条件当遍历到这个节点时权值已经大于等于w但它不是叶节点。不过估计这题不剪枝也没啥
问题。。。牛客网能过，pat第三个测试例子死活过不了，求大佬指教。。。#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct TNode{
int weight;
vector<int> child;
};
TNode tree[100];
int w;
vector<int> tmp;
vector<vector<int>> ans;
void dfs(int rt){
static int sum = 0;
sum += tree[rt].weight;
tmp.push_back(tree[rt].weight);
if(tree[rt].child.empty()&&sum==w)
ans.push_back(tmp);
else if(!tree[rt].child.empty()&&sum>=w){
tmp.pop_back();
sum -= tree[rt].weight;
return;
}
for(int i=0;i<tree[rt].child.size();++i)
dfs(tree[rt].child[i]);
tmp.pop_back();
sum -= tree[rt].weight;
}
bool cmp(vector<int> a,vector<int> b){
for(int i=0;i<a.size()&&i<b.size();++i)
if(a[i]>b[i])
return 1;
return a.size()>b.size();
}
int main()
{
int n,m,id,k,num;
cin>>n>>m>>w;
for(int i=0;i<n;++i)
cin>>tree[i].weight;
for(int i=0;i<m;++i){
cin>>id>>k;
while(k--){
cin>>num;
tree[id].child.push_back(num);
}
}
dfs(0);
sort(ans.begin(),ans.end(),cmp);
for(auto i:ans){
cout<<i[0];
for(int j=1;j<i.size();++j)
cout<<' '<<i[j];
cout<<endl;
}
return 0;
}