1053. Path of Equal Weight (30)

题目：

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 Figure 1: 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 Figure 1.

Figure 1

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 W_i (<1000) corresponds to the tree node T_i. 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, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, and A_k+1 > B_k+1.

Sample Input:

20 9 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

注意：

1、dfs遍历，遍历所有路径，计算出每个路径的长度（注意每条路径必须是从根节点到无叶节点，不能是中间的一段）。

2、由于要从小到大输出，所以在输入的时候对每一个结点的所有子结点排序，这样dfs的时候就是按照从小到大的顺序来遍历的，这样直接输出就可以了。

代码：

//1053#include<iostream>#include<vector>#include<algorithm>using namespace std;vector<vector<int>>child;//child nodes of each nodevector<int>weight;//weight of each nodevector<int>pre;//previous node of each nodevector<vector<int>>ans;//sequence with sum weight equals to svector<int>visited;//flags whether visitedvector<int>sum;//sum weight of each path from root to each nodeint n,m,s;vector<int> path(int n){vector<int>a,b;while(n!=0){a.push_back(weight[n]);n=pre[n];}b.push_back(weight[0]);for(int i=a.size()-1;i>=0;--i)b.push_back(a[i]);return b;}bool cmp(int a,int b){return weight[a]>weight[b];}void dfs(int n){visited[n]=1;for(int i=0;i<child[n].size();++i)if(!visited[child[n][i]]){sum[child[n][i]] = sum[n]+weight[child[n][i]];if(sum[child[n][i]]==s && child[child[n][i]].empty())ans.push_back(path(child[n][i]));dfs(child[n][i]);}}int main(){scanf("%d%d%d",&n,&m,&s);weight.resize(n);pre.resize(n);for(int i=0;i<n;++i)scanf("%d",&weight[i]);child.resize(n);for(int i=0;i<m;++i){int id,k;scanf("%d%d",&id,&k);child[id].resize(k);for(int j=0;j<k;++j){scanf("%d",&child[id][j]);pre[child[id][j]]=id;}//sort by weight so that the ans vector will be sorted atomaticallysort(child[id].begin(),child[id].end(),cmp);}visited.assign(n,0);sum.assign(n,0);sum[0]=weight[0];if(weight[0]==s){//case 1printf("%d",weight[0]);return 0;}for(int i=0;i<n;++i){if(!visited[i])dfs(i);}for(int i=0;i<ans.size();++i){for(int j=0;j<ans[i].size()-1;++j)printf("%d ",ans[i][j]);printf("%d\n",ans[i][ans[i].size()-1]);}return 0;}