【PAT】【Advanced Level】1053. Path of Equal Weight (30)

1053. Path of Equal Weight (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 Lis 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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1053

https://www.nowcoder.com/pat/5/problem/4092

思路：

DFS+条件判断

CODE：

#include<iostream>#include<cstring>#include<vector>#include<algorithm>#define N 101using namespace std;vector<int> v[N];int now[N];int c[N];int zs=0;int maxx;int fl;bool cmp(int a,int b){return c[a]>c[b];}void dfs(int n,int valu){if (v[n].size()==0){if (valu==maxx) { for (int i=0;i<fl-1;i++){cout<<c[now[i]]<<" ";}cout<<c[now[fl-1]]<<endl;zs++; }return;}for (int i=0;i<v[n].size();i++){now[fl]=(v[n][i]);fl++;dfs(v[n][i],valu+c[v[n][i]]);fl--;}return;}int main(){int n,m,va;cin>>n>>m>>maxx;for (int i=0;i<n;i++){cin>>c[i];}for (int i=0;i<m;i++){int no;cin>>no;int num;cin>>num;for (int j=0;j<num;j++){int nod;cin>>nod;v[no].push_back(nod);}sort(v[no].begin(),v[no].end(),cmp);}fl=0;now[fl]=0;fl++;dfs(0,c[0]);return 0;}