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: 给你一棵树,树上的每个结点有特定的权值,要求找到这么一条路径,起点是根节点,终点是叶子结点,路径上各结点的权值之和等于S
思路:输出要求类似于字符串的字典序,那么可以将每个结点的子节点按权值从大到小排序,这样在遍历时输出自然符合要求
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const int Max=105;
int Wei[Max],Mark[Max],Num;
int N,M,S;
vector<int> node[Max];
vector<int> Temp;
bool Compare(int x,int y)
{
return Wei[x]>Wei[y];
}
void DFS(int cur,LL v)
{
if(v>S)
return ;
if(v==S)
{
if(!node[cur].size())
{
int Len=Temp.size();
for(int i=0;i<Len;i++)
{
if(i==0)
printf("%d",Temp[i]);
else
printf(" %d",Temp[i]);
}
printf("\n");
}
return ;
}
int Len=node[cur].size();
for(int i=0;i<Len;i++)
{
int x=node[cur][i];
if(!Mark[x])
{
Mark[x]=1;
Temp.push_back(Wei[x]);
DFS(x,v+Wei[x]);
Mark[x]=0;
Temp.pop_back();
}
}
return ;
}
int main()
{
scanf("%d %d %d",&N,&M,&S);
memset(Mark,0,sizeof(Mark));
for(int i=0;i<N;i++)
{
scanf("%d",&Wei[i]);
}
int x,y,z;
while(M--)
{
scanf("%d %d",&x,&y);
for(int i=0;i<y;i++)
{
scanf("%d",&z);
node[x].push_back(z);
}
sort(node[x].begin(),node[x].end(),Compare);
}
Num=0;
Temp.push_back(Wei[0]);
Mark[0]=1;
DFS(0,Wei[0]);
return 0;
}