1053. Path of Equal Weight (30)【树+搜索】——PAT (Advanced Level) Practise

题目信息

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

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. 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 < 2^30, the given weight number. The next line contains N positive numbers where Wi (<1000) 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 {A1, A2, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, … k, 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

解题思路

建树然后深搜

AC代码

#include <cstdio>
#include <vector>
#include <algorithm>
#include <functional>
using namespace std;
int a[105], r[105];
int n, m, w, t, id, d;
vector<int> node[105];
bool cmp(const int v1, const int v2){
return a[v1] > a[v2];
}
void dfs(int root, int lv, int wt){
r[lv] = a[root];
wt += a[root];
if (node[root].empty() && wt == w){
printf("%d", r[0]);
for (int i = 1; i <= lv; ++i){
printf(" %d", r[i]);
}
printf("\n");
return;
}
sort(node[root].begin(), node[root].end(), cmp);
for (int i = 0; i < node[root].size(); ++i){
dfs(node[root][i], lv + 1, wt);
}
}
int main()
{
scanf("%d%d%d", &n, &m, &w);
for (int i = 0; i < n; ++i){
scanf("%d", a+i);
}
for (int i = 0; i < m; ++i){
scanf("%d%d", &id, &t);
for (int j = 0; j < t; ++j){
scanf("%d", &d);
node[id].push_back(d);
}
}
dfs(0, 0, 0);
return 0;
}