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 Rto 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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains Npositive numbers where Wi (<) 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 { is said to be greater than sequence { if there exists 1 such that Ai=Bi for ,, 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

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

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-21-15.57.11
 * Description : A1053
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 struct node{
     int weight;
     vector<int> child;
 }Node[maxn];
 int n,m,s;
 vector<int> path;
 bool cmp(int a,int b){
     return Node[a].weight>Node[b].weight;
 }
 //bool isUsed[maxn]={false};
 void DFS(int index,int sum){
     if(index>=n||sum>s) return;
     if(sum==s){
         ) return;
         int len=path.size();
         ;i<len;i++){
             printf("%d",Node[path[i]].weight);
             ) printf(" ");
             else printf("\n");
         }
         return;
     }
     int t=Node[index].child.size();
     ;i<t;i++){
         int child=Node[index].child[i];
         //if(isUsed[child]==true) continue;
         path.push_back(child);
         //isUsed[child]=true;
         DFS(child,sum+Node[child].weight);
         //isUsed[child]=false;
         path.pop_back();
     }
 }

 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int temp,k,c;
     scanf("%d%d%d",&n,&m,&s);
     ;i<n;i++){
         scanf("%d",&Node[i].weight);
     }
     ;i<m;i++){
         scanf("%d%d",&temp,&k);
         ;j<k;j++){
             scanf("%d",&c);
             Node[temp].child.push_back(c);
         }
         //sort(Node[i].child.begin(),Node[i].child.end(),cmp);
         sort(Node[temp].child.begin(),Node[temp].child.end(),cmp);
     }
     path.push_back();
     DFS(,Node[].weight);
     ;
 }

巴特西

1053 Path of Equal Weight （30 分）

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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