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题目大意：给出一棵树，每个节点有一个编号和一个权重，输入给定了一个总权重K，求根节点R到任一叶子节点的路径代价=总权重K，找出所有这样的路径，并按字典序从大到小输出。
//我的代码：写不下去的：
#include <cstdio>
#include<iostream>
#include <vector>
#include<set>
#include<map>
using namespace std;
vector<int> vt[];
map<int,int> mp;
vector<int> path;//用什么去存这个路径呢？
set<vector<int>> st;
void dfs(int r){//你这sum都没有作为参数传进去欸。。。
if(vt[r].size()==){//这里你还判断错了。。
for(int i=;i<path.size();i++){
st.insert(path);
}
return ;
}
path.push_back(r);
//最后怎么将其按照字典序排序呢？
for(int i=vt[r][];i<vt[r].size();i++){
path.push_back(vt[r][i]);
dfs(i);//.没有这个函数啊.
path.erase(vt[r][i]);
} }
int main(){
int n,leaf,k;
cin>>n>>leaf>>k;
int weight;
for(int i=;i<n;i++){
cin>>weight;
mp[i]=weight;
}
int nei=n-leaf;
int u,ct,temp;
for(int i=;i<nei;i++){
cin>>u>>ct;
for(int j=;j<ct;j++){
cin>>temp;
vt[u].push_back(temp);
}
}
//path.push_back(0);
dfs();
return ;
}
//dfs很多东西都没考虑好。
代码来自：https://www.liuchuo.net/archives/2285
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int target;
struct NODE {
int w;
vector<int> child;//将孩子节点作为一个向量进行存储。便于排序
};
vector<NODE> v;
vector<int> path;
void dfs(int index, int nodeNum, int sum) {
if(sum > target) return ;
if(sum == target) {
if(v[index].child.size() != ) return;//如果不是叶节点那么也返回。
for(int i = ; i < nodeNum; i++)
printf("%d%c", v[path[i]].w, i != nodeNum - ? ' ' : '\n');//直接在输出判断，十分简洁。
return ;
}
for(int i = ; i < v[index].child.size(); i++) {
int node = v[index].child[i];
path[nodeNum] = node;//不是push_back，向量没有erase函数，所以传了个参数表示节点数量。
dfs(node, nodeNum + , sum + v[node].w);
} }
int cmp1(int a, int b) {
return v[a].w > v[b].w;
}
int main() {
int n, m, node, k;
scanf("%d %d %d", &n, &m, &target);
v.resize(n), path.resize(n);
for(int i = ; i < n; i++)
scanf("%d", &v[i].w);
for(int i = ; i < m; i++) {
scanf("%d %d", &node, &k);
v[node].child.resize(k);
for(int j = ; j < k; j++)
scanf("%d", &v[node].child[j]);
sort(v[node].child.begin(), v[node].child.end(), cmp1);
//对子节点从大到小排序，这样就能保证是按字典序最大来找到并输出的。
}
dfs(, , v[].w);
return ;
}
1.如何保证输出的路径是按字典序从大到小呢？将每个节点的子节点按权重从大到小排列即可。学习了
2.path如果用Vector表示，但是又不能弹出，该怎么办呢？在dfs中传入参数nodeNum,而不是在跳出递归进行计算时，使用path.size()；下一次的就被覆盖了。
3.要注意dfs传入的参数，当前下标、解中节点数、总和。
//值得学习！