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
题意:
给定一棵树和每个结点的权值,求所有从根结点到叶子结点的路径,使得每条路径上的结点的权值之和等于给定的常数S(按顺序输出路径上结点的权值)。如果有多条这样的路径,按路径非递增的顺序输出。
分析:
①因为并非二叉树,所以子树不止2棵;因为题目给出每个结点编号,所以采用数组静态构建树更为方便,直接用编号做为下标(地址)。
struct node
{
int W; //权值
vector<int> sub; //用于存储子树编号
}Node[110];
②关于最后要按路径非递增输出,用vector就可以很方便的处理,因为vector<基本数据类型>是可以直接用关系运算符(>,<,<=,>=,!=,==)进行比较的,那么就可以构造cmp函数了。
vector<vector<int> > ans;
bool cmp(vector<int> a,vector<int> b )
{
return a>b;
}
以下完整代码:
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node
{
int W;
vector<int> sub;
}Node[110];
int M,N,S;
vector<int> temp;
vector<vector<int> > ans;
bool cmp(vector<int> a,vector<int> b )
{
return a>b;
}
void DFS(int root,int sum)
{
if(sum>=S)
{
if(sum==S&&Node[root].sub.empty()) //后一个判断条件确保结束结点为叶结点
ans.push_back(temp); //叶结点无子树,其sub为空
return;
}
for(int i=0;i<Node[root].sub.size();i++)
{
int next=Node[root].sub[i];
temp.push_back(Node[next].W);
DFS(next,sum+Node[next].W);
temp.pop_back();
}
}
int main()
{
scanf("%d %d %d",&N,&M,&S);
for(int i=0;i<N;i++)
scanf("%d",&Node[i].W);
for(int i=0;i<M;i++)
{
int ID,K,t;
scanf("%d %d",&ID,&K);
while(K--)
{
scanf("%d",&t);
Node[ID].sub.push_back(t);
}
}
temp.push_back(Node[0].W); //先把根结点放入
DFS(0,Node[0].W);
sort(ans.begin(),ans.end(),cmp); //排序
for(int i=0;i<ans.size();i++) //输出
{
for(int j=0;j<ans[i].size();j++)
{
if(j!=0)
printf(" ");
printf("%d",ans[i][j]);
}
printf("\n");
}
return 0;
}