1053 Path of Equal Weight （30 分）

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原文链接：http://www.cnblogs.com/Mered1th/p/10413731.html

本文介绍了一个算法问题，即在一个加权树中寻找所有路径，这些路径的权重之和等于给定数值。通过深度优先搜索（DFS）遍历树并回溯，文章详细描述了解决方案的实现过程，包括输入输出规格及示例。

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Given a non-empty tree with root

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 $0, the number of nodes in a tree,$

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 ${ is said to be greater than sequence { if there exists 1 such that$

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



分析：DFS 30分里的水题。。回溯一遍就行了。。但测试点2一直过不了。。

2.24更新，发现问题了，sort排序的时候应该对temp排序，太粗心了。另外isused数组也没啥用。。删了即可

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-21-15.57.11
 6 * Description : A1053
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=110;
20 struct node{
21     int weight;
22     vector<int> child;
23 }Node[maxn];
24 int n,m,s;
25 vector<int> path;
26 bool cmp(int a,int b){
27     return Node[a].weight>Node[b].weight;
28 }
29 //bool isUsed[maxn]={false};
30 void DFS(int index,int sum){
31     if(index>=n||sum>s) return;
32     if(sum==s){
33         if(Node[index].child.size()!=0) return;
34         int len=path.size();
35         for(int i=0;i<len;i++){
36             printf("%d",Node[path[i]].weight);
37             if(i!=len-1) printf(" ");
38             else printf("\n");
39         }
40         return;
41     }
42     int t=Node[index].child.size();
43     for(int i=0;i<t;i++){
44         int child=Node[index].child[i];
45         //if(isUsed[child]==true) continue;
46         path.push_back(child);
47         //isUsed[child]=true;
48         DFS(child,sum+Node[child].weight);
49         //isUsed[child]=false;
50         path.pop_back();
51     }
52 }
53 
54 int main(){
55 #ifdef ONLINE_JUDGE
56 #else
57     freopen("1.txt", "r", stdin);
58 #endif
59     int temp,k,c;
60     scanf("%d%d%d",&n,&m,&s);
61     for(int i=0;i<n;i++){
62         scanf("%d",&Node[i].weight);
63     }
64     for(int i=0;i<m;i++){
65         scanf("%d%d",&temp,&k);
66         for(int j=0;j<k;j++){
67             scanf("%d",&c);
68             Node[temp].child.push_back(c);
69         }
70         //sort(Node[i].child.begin(),Node[i].child.end(),cmp);
71         sort(Node[temp].child.begin(),Node[temp].child.end(),cmp);
72     }
73     path.push_back(0);
74     DFS(0,Node[0].weight);
75     return 0;
76 }

转载于:https://www.cnblogs.com/Mered1th/p/10413731.html