PAT (Advanced Level) Practise 1053 Path of Equal Weight (30)

1053. Path of Equal Weight (30)

时间限制

10 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 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_i for 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 2
10 3 3 6 2
统计并输出从根节点到叶子节点的路径中和是s的路径
#include<cmath>#include<vector>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int mod = 1e9 + 7;const int maxn = 2e5 + 10;int T, n, m, s, x, y, z, v[maxn], u[maxn], tot;vector<int> t[maxn], ans[maxn];void dfs(int x, int y, int dep){u[dep] = v[x];y += v[x];if (t[x].size() == 0 && y == s){for (int i = 0; i <= dep; i++){ans[tot].push_back(u[i]);}tot++;}for (int i = 0; i < t[x].size(); i++){dfs(t[x][i], y, dep + 1);}}bool cmp(const vector<int>&a, const vector<int>&b){for (int i = 0; i < min(a.size(), b.size()); i++){if (a[i] == b[i]) continue;return a[i] > b[i];}return false;}int main(){scanf("%d%d%d", &n, &m, &s);for (int i = 0; i < n; i++) scanf("%d", &v[i]);for (int i = 0; i < m; i++){scanf("%d%d", &x, &y);while (y--) scanf("%d", &z), t[x].push_back(z);}dfs(0, 0, 0);sort(ans, ans + tot, cmp);for (int i = 0; i < tot; i++){for (int j = 0; j < ans[i].size(); j++){if (j) printf(" ");printf("%d", ans[i][j]);}printf("\n");}return 0;}