1053. Path of Equal Weight (30) DFS

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 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 < 230, 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 = Bifor 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

题意：给定一棵树，求根节点到叶节点的权值和为定值的路径

题解 : DFS

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 110;
int s;
int weight
, ary
;

struct node{
vector<int> v;
bool operator<(const node& X) const{
return v > X.v;
}
};
vector<node> ans;
vector<int> vertex
;

void solve(int cur_node, int cur_num, int total) {
if (total > s) return ;
if (vertex[cur_node].size() == 0 && total == s) {
vector<int> v;
for (int i = 0; i < cur_num; ++i)
v.push_back(weight[ary[i]]);
node tmp = {v};
ans.push_back(tmp);
return ;
}
for (int i = 0; i < vertex[cur_node].size(); ++i) {
ary[cur_num] = vertex[cur_node][i];
solve(vertex[cur_node][i], cur_num + 1, total + weight[vertex[cur_node][i]]);
}
}

int main() {
int n, m;
scanf("%d%d%d", &n, &m, &s);
for (int i = 0; i < n; ++i)
scanf("%d", weight + i);

for (int i = 0; i < m; ++i) {
int id, num, tmp;
scanf("%d%d", &id, &num);
while (num--) {
scanf("%d", &tmp);
vertex[id].push_back(tmp);
}
}

solve(0, 1, weight[0]);
sort(ans.begin(), ans.end());
for (int i = 0; i < ans.size(); ++i) {
for (int j = 0; j < ans[i].v.size(); ++j) {
printf("%d%c", ans[i].v[j], j == ans[i].v.size() - 1 ? '\n' : ' ');
}
}

return 0;
}