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 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 = Bifor 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 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() == &&add==sum) {//符合条件
tmp.clear();
for (int i = ; i <=step; i++)tmp.push_back(road[i]);
r.push_back(tmp);
return;
}
for (int i = ; i < G[x].size();i++) {
Node p = G[x][i];
dfs(p.id, step + , p.key + add);
}
} bool cmp(vector<int>a,vector<int>b ) {
int i = ;
while (a[i] == b[i]&&i<a.size()-&&i<b.size()-)i++;
return a[i] > b[i];
} int main() {
while (cin>>n>>m>>sum) {
for (int i = ; i < n; i++) {
int a; cin >> a;
node[i] = Node(i, a);
}
for (int i = ; i < m;i++) {
int from, k;
cin >> from >> k;
while (k--) {
int to; cin >> to;
G[from].push_back(node[to]);
}
}
dfs(, , node[].key);
sort(r.begin(), r.end(), cmp);
for (int i = ; i < r.size();i++) {
for (int j = ; j < r[i].size();j++) {
printf("%d%c",r[i][j],j+==r[i].size()?'\n':' ');
}
}
}
return ;
}

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