1053 Path of Equal Weight (30 分)
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
accept code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int N,M,S;
int w[110];
int path[110];//用来记录路径的结点
struct Node{
int weight;
vector<int> child;
}node[110];
bool cmp(int a,int b)
{
return node[a].weight>node[b].weight;
}
//index是当前访问结点,numnode是当前路径path上的结点的个数,sum是当前的结点的权值之和
void presort(int index, int numnode,int sum)//先根遍历所有的结点
{
if(sum>S) return;//超过了给定的权重和
if(sum==S)//刚好相等做相应处理
{
if(node[index].child.size()!=0) return;//证明不是叶子结点
for(int q=0;q<numnode;q++)//否则输出当前路径的结点的权重
{
printf("%d",node[path[q]].weight);
if(q<numnode-1) printf(" ");
else printf("\n");
}
return;
}
//遍历当前结点的所有子节点
for(int r=0;r<node[index].child.size();r++)
{
int child=node[index].child[r];//子节点标号
path[numnode]=child;
presort(child,numnode+1,sum+node[child].weight);
}
}
int main()
{
cin>>N>>M>>S;
for(int i=0;i<N;i++)
cin>>node[i].weight;//存放每个节点的权重,下标从0开始且每个两位数的节点(下标)对应的值就是相应权重
//接下来有M行,每一行一个结点以及其权重信息和其子节点的标号
int t;//该结点的子节点个数
int nod;//该结点的下标
int nextnod;
for(int j=0;j<M;j++)
{
// cin>>nod>>t;
scanf("%d%d",&nod,&t);
for(int k=0;k<t;k++)
{
// cin>>nextnod;
scanf("%d",&nextnod);
// node[t].data=w[nod];
node[nod].child.push_back(nextnod);
}
//将每个节点所连接的子节点的权重进行排序,从大到小
sort(node[nod].child.begin(),node[nod].child.end(),cmp);
}
path[0]=0;
presort(0,1,node[0].weight);
return 0;
}