PAT日志 1053

顽强的小白

1053 Path of Equal Weight （30 分）

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. 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.

题目解析

找到路径权值之和为题目要求的路径

代码实现

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=105;
struct {
 int weight;
 vector<int> child;
}node[maxn];
int n,m,s;
vector<int> tmp;    //储存 路径结点信息 
bool cmp(int a,int b){
 return node[a].weight>node[b].weight;
}
void DFS(int index,int sum){
 if(sum>s) return;          //这条路径权值太大了 
 if(sum==s){                   //满足和的要求 
  if(node[index].child.size()==0){  //还要判断是不是到叶结点了 
   for(int i=0;i<tmp.size();++i){     //直接在这里输出 
    printf("%d",node[tmp[i]].weight);
    if(i<tmp.size()-1) printf(" ");
    else printf("\n");
   }
  }
  return;
 }
 
 for(int i=0;i<node[index].child.size();++i){   //遍历所有孩子 
  int child=node[index].child[i];
  tmp.push_back(child);                     //走这个孩子的分支 
  DFS(child,sum+node[child].weight); 
  tmp.pop_back();                         //走完这个分支就退回来 
 }
}
int main(){
 scanf("%d%d%d",&n,&m,&s);
 for(int i=0;i<n;++i){
  scanf("%d",&node[i].weight);
 }
 int k,child,now;
 for(int i=0;i<m;++i){
  scanf("%d%d",&now,&k);
  for(int j=0;j<k;++j){
   scanf("%d",&child);
   node[now].child.push_back(child);
  }
  sort(node[now].child.begin(),node[now].child.end(),cmp);     
  //一定要记得要把孩子们按照权值从大到小排序 
  //不然就会按照编号大小，输出的结果不满足要求 
 }                                                              
 
 tmp.push_back(0);         //添上根结点 
 DFS(0,node[0].weight) ;   //根节点的权值也要有 
 return 0;
}