A1053 Path of Equal Weight (30 point(s))

DFS 取和为s的路径

1. 原文

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.

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 {$A_1$,$A_2$,⋯,$A_n$} is said to be greater than sequence {$B_1$$B_2$,⋯,$B_m$} if there exists 1≤k<min{n,m}such that $A_i$=$B_i$ for *i*=1,⋯,*k*, and $A_{k+1}$>$B_{k+1}$.

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

2. 解析

按权重从高到低排序，return v[a].weight>v[b].weight;

和为s if (sum==s)

记录路径 path[cnt]=idx;

dfs(temp, sum+v[temp].weight,index+1);

• 最后一个测试点错误❌：

  7 5 8
  1 2 2 2 3 3 2
  00 2 01 02
  01 1 03
  02 1 04
  03 1 05
  04 1 06

  期待的结果是

  1 2 3 2

  1 2 2 3

  实际的结果是

  1 2 2 3

  1 2 3 2

• 因为这种判断逻辑保证的是在同一层结点权值不同时，选择权值最大的结点。但是同一层有权值相同的结点时，选择的结点下一层孩子结点权值可能不是该同层最大的

• 需要对结果也进行排序，greater<T>在排序中是将数组元素从大到小排序

3. AC代码

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
    int weight;
    vector<int> child;
};
vector<int> path;
vector<node> Nodes;
vector<vector<int>> ans;
int n,m,s;
bool cmp(int a,int b){
    return Nodes[a].weight>Nodes[b].weight;
}
void dfs(int index,int sum){
    if(sum>s){
        return;
    }
    if(sum==s){
        if(Nodes[index].child.size()!=0){
            return;
        }
        ans.push_back(path);
        return;
    }
    for(int i=0;i<Nodes[index].child.size();i++){
        int idx = Nodes[index].child[i];
        path.push_back(Nodes[idx].weight);
        dfs(idx,sum+Nodes[idx].weight);
        path.pop_back();
    }
    
}
int main(){
    scanf("%d%d%d",&n,&m,&s);
    Nodes.resize(n);
    for(int i=0;i<n;i++){
        scanf("%d",&Nodes[i].weight);
    }
    for(int i=0;i<m;i++){
        int a,k,b;
        scanf("%d%d",&a,&k);
        Nodes[a].child.resize(k);
        for(int j=0;j<k;j++){
            scanf("%d",&Nodes[a].child[j]);
        }
        sort(Nodes[a].child.begin(),Nodes[a].child.end(),cmp);
    }
    path.push_back(Nodes[0].weight);
    dfs(0,Nodes[0].weight);
    sort(ans.begin(),ans.end(),greater<vector<int>>());
    for(int i=0;i<ans.size();i++){
        for(int j=0;j<ans[i].size();j++){
            printf("%d",ans[i][j]);
            if(j!=ans[i].size()-1){
                printf(" ");
            }else{
                printf("\n");
            }
        }
    }
    return 0;
}