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 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 = 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

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;
}