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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

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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;}