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 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 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 101 using 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; }