PAT 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 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

---

提交代码

深搜即可，但是要求输出的路径是按路径的序列，可以用优先队列

讲能到达的点放进去，再深搜。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n, m, p, k, ans, a[maxn], head[maxn], path[maxn];
struct node
{
    int v, w, next;
}edge[maxn];

struct node2
{
    int v, w;
    node2() {}
    node2(int vv, int ww): v(vv), w(ww) {}
    bool operator < (const node2 &a) const
    {
        return w < a.w;
    }
};

void addEdge(int u, int v)
{
    edge[k].v = v;
    edge[k].next = head[u];
    head[u] = k++;
}

void dfs(int cur, int setp, int sum)
{
    if(head[cur] == -1 && sum == p)
    {
        for(int i = 1; i < setp; i++)
        {
            if(i-1) printf(" ");
            printf("%d", path[i]);
        }
        printf("\n");
    }
    priority_queue<node2> pq;
    for(int u = head[cur]; u != -1; u = edge[u].next)
    {
        int v = edge[u].v;
        pq.push(node2(v, a[v]));
    }
    while(!pq.empty())
    {
        node2 t = pq.top(); pq.pop();
        path[setp] = t.w;
        dfs(t.v, setp+1, sum+t.w);
    }
}

int main(void)
{
    while(cin >> n >> m >> p)
    {
        k = 0;
        memset(head, -1, sizeof(head));
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= m; i++)
        {
            int u, v, num;
            scanf("%d%d", &u, &num);
            while(num--)
            {
                scanf("%d", &v);
                addEdge(u+1, v+1);
            }
        }
        path[1] = a[1];
        dfs(1, 2, a[1]);
    }
    return 0;
}