PAT 1053. Path of Equal Weight

寻找等权路径

最新推荐文章于 2022-05-13 18:14:30 发布

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本文介绍了一个寻找树中所有等权路径的问题，通过DFS遍历树结构，并利用C++ STL库进行辅助实现。针对给定权重，算法能找出所有路径并按非递增顺序输出。

思路就是：

先用map把不是根节点的子节点存储起来，并排序，然后 DFS遍历，同时用一个vector保存路径，当遍历到叶子节点，并且加起来正好满足条件时就print。主要是C++的STL库不熟，加上排序定义比较器的意识也没有，所以有点无从下手的感觉。所以主要参考了别人的博客

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:

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

using namespace std;

int N, M, sum, f, n, c;
int weight[102];
map<int, vector<int> > adjlist;
vector<int> trace;

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

void dfs(int start, int cur_sum) {
    if(cur_sum > sum)   return;

    if(sum == cur_sum && adjlist[start].empty()) {
        for(int i=0; i<trace.size()-1; i++)
            printf("%d ", weight[trace[i]]);
        printf("%d\n", weight[trace[trace.size()-1]]);
    }

    // may throw exception

    for(int i=0; i<adjlist[start].size(); i++) {
        int next = adjlist[start][i];
        trace.push_back(next);
        dfs(next, cur_sum + weight[next]);
        trace.pop_back();
    }
}

int main()
{
    scanf("%d %d %d", &N, &M, &sum);
    for(int i=0; i<N; i++)
        scanf("%d", &weight[i]);
    for(int i=0; i<M; i++) {
        scanf("%d %d", &f, &n);
        vector<int> v;
        for(int j=0; j<n; j++) {
            scanf("%d", &c);
            adjlist[f].push_back(c);
        }
        sort(adjlist[f].begin(), adjlist[f].end(), cmp);
    }

    trace.push_back(0);
    dfs(0, weight[0]);
    return 0;
}