1053. Path of Equal Weight (30)
2015-12-06 11:06
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dfs函数携带vector形参记录搜索路径
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 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 1Input 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:
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:
[/code]
来自为知笔记(Wiz)
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 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 1Input 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 19Sample Output:
10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2来源: <http://www.patest.cn/contests/pat-a-practise/1053>
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
map<int, vector<int>> mp;
vector<int> weight;
int n, m, aimWeight;
int pathWeight=0;
bool cmp(int a, int b) {
return weight[a] > weight[b];
}
void dfs(int i,int w, vector<int> p) {
p.push_back(i);
if (mp[i].size() == 0) {
if (w == aimWeight) {
for (int j = 0; j < p.size(); j++) {
cout << weight[p[j]];
if (j != p.size() - 1)
cout << " ";
else
cout << endl;
}
}
p.clear();
return;
}
else {
for (int j = 0; j < mp[i].size(); j++)
dfs(mp[i][j],weight[mp[i][j]]+w,p);
}
}
int main(void) {
cin >> n >> m >> aimWeight;
for (int i = 0; i < n; i++) {
int wtemp;
cin >> wtemp;
weight.push_back(wtemp);
}
for (int i = 0; i < m; i++) {
int id, k;
cin >> id >> k;
for (int j = 0; j < k; j++) {
int idtemp;
cin >> idtemp;
mp[id].push_back(idtemp);
}
sort(mp[id].begin(), mp[id].end(), cmp);
}
vector<int> path;
dfs(0,weight[0],path);
return 0;
}
[/code]
来自为知笔记(Wiz)
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