pat 甲级 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 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_ifor 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

题意：寻找所有节点键值的和为指定值的路径，并且按照一定规则对路径排序后输出。
思路：dfs
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 100+5
typedef long long ll;
int n, m, sum;
struct Node {
    int key,id;
    Node() {}
    Node(int id,int key):id(id),key(key) {}
}node[N_MAX];
vector<Node>G[N_MAX];

int road[N_MAX];
vector<vector<int> >r; 
vector<int>tmp;
void dfs(int x,int step,int add) {
    road[step] = node[x].key;
    if (G[x].size() == 0 &&add==sum) {//符合条件
        tmp.clear();
        for (int i = 0; i <=step; i++)tmp.push_back(road[i]);
        r.push_back(tmp);
        return;
    }
    for (int i = 0; i < G[x].size();i++) {
        Node p = G[x][i];
        dfs(p.id, step + 1, p.key + add);
    }
}

bool cmp(vector<int>a,vector<int>b ) {
    int i = 0;
    while (a[i] == b[i]&&i<a.size()-1&&i<b.size()-1)i++;
    return   a[i] > b[i];
}


int main() {
    while (cin>>n>>m>>sum) {
        for (int i = 0; i < n; i++) {
            int  a; cin >> a;
            node[i] = Node(i, a);
        }
        for (int i = 0; i < m;i++) {
            int from, k;
            cin >> from >> k;
            while (k--) {
                int to; cin >> to;
                G[from].push_back(node[to]);
            }
        }
        dfs(0, 0, node[0].key);
        sort(r.begin(), r.end(), cmp);
        for (int i = 0; i < r.size();i++) {
            for (int j = 0; j < r[i].size();j++) {
                printf("%d%c",r[i][j],j+1==r[i].size()?'
':' ');
            }
        }
    }
    return 0;
}