PAT-A-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 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

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;
const int maxn = 110;
int n, m, sum;
struct node
{
int weight;
vector<int> child;

}Node[maxn];

int cmp(int a, int b)
{
return Node[a].weight > Node[b].weight;
}

int path[maxn];
void DFS(int index, int nodenum, int nowsum)
{
if (nowsum > sum)
return;
if (nowsum == sum)
{
if (Node[index].child.size() == 0)
{
for (int i = 0; i < nodenum; i++)
{
cout << path[i];
if (i < nodenum - 1)
cout << " ";
}

cout << endl;
}

return;
}

for (int i = 0; i < Node[index].child.size(); i++)
{
path[nodenum] = Node[Node[index].child[i]].weight;
DFS(Node[index].child[i], nodenum+1, nowsum + Node[Node[index].child[i]].weight);
}
}

int main()
{
cin >> n >> m >> sum;

for (int i = 0; i < n; i++)
{
cin >> Node[i].weight;
}

int num, fu, son;
for (int i = 0; i < m; i++)
{
cin >> fu >> num;
for (int j = 0; j < num; j++)
{
cin >> son;
Node[fu].child.push_back(son);
}

sort(Node[fu].child.begin(), Node[fu].child.end(), cmp);

}

path[0] = Node[0].weight;
DFS(0, 1, Node[0].weight);

system("pause");
return 0;
}