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
Special thanks to Zhang Yuan and Yang Han for their contribution to the judge’s data.
题意:给定一棵树,从根节点开始每一个节点都有它的权重,从根节点开始搜索权重和为S的节点,若搜索到则·依次输出他们各自的权重,若有相同者,则从大到小依次输出。
思路:进行DFS搜索,因为要从大到小依次输出,所以要为每一层的节点进行从大到小的排序!!!
AC
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=101;
struct node{
int weight;
vector<int>child;
}Node[maxn];//树的静态结构
bool cmp(int a,int b){//排序判断
return Node[a].weight>Node[b].weight;
}
int n,m,s;
vector<int>vc;
void DFS(int root,int sum){
if(Node[root].child.size()==0){//判断是否为叶子节点
if(sum==s){//权重等于
for(int i=0;i<vc.size();i++){//格式输出
if(i<vc.size()-1){
printf("%d ",vc[i]);
}else{
printf("%d\n",vc[i]);
}
}
}
return;
}
for(int i=0;i<Node[root].child.size();i++){//孩子结点的DFS
int child=Node[root].child[i];
vc.push_back(Node[child].weight);//选择第i个节点进入数组
DFS(child,sum+Node[child].weight);
vc.pop_back();//结束第i个节点
}
}
int main(){
scanf("%d%d%d",&n,&m,&s);
for(int i=0;i<n;i++){
scanf("%d",&Node[i]);
}
int id,k,child;
for(int i=0;i<m;i++){
scanf("%d%d",&id,&k);
for(int j=0;j<k;j++){
scanf("%d",&child);
Node[id].child.push_back(child);
}
sort(Node[id].child.begin(),Node[id].child.end(),cmp);//每层节点排序
}
vc.push_back(Node[0].weight);
DFS(0,Node[0].weight);
}
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