華爲軟件精英挑戰賽2020題目
爲了方便理解題目,暴力求解了一下。。。
每個點都進行深度遍歷,找到以這個點爲起點的環。。。數據再大一點應該就不行了。。。
代碼寫得爛,僅供參考。。
更新,用鄰接表來實現速度快了100倍。。。
更新,根據大佬的Java代碼寫了兩個小的數據集。。。
https://github.com/izhangrui/HWcode2020-TestData/tree/master/C%2B%2B
更新,數據連通性太強了,拓撲排序只能去除一兩個點,看情況使用。。。。
同一份代碼,爲什麼用vs跑要比g++快?
100w的數據排序用太長時間了。。。是不是可以考慮排序用多線程。。。
更新,g++慢是因爲編譯的時候沒加優化選項-O3。。。
試了一下鯤鵬服務器,感覺和筆記本差不多。。。
看這個100w和300w的結果,這用時和環數呈正比?
另外排序慢很有可能是因爲生成的數據高度有序?快排貌似對有序的數據很慢。。。
更新,貌似可以省去對結果排序。。。
#include "bits/stdc++.h"
using namespace std;
void dfs(const vector<vector<int>> &g, vector<vector<int>> &res, vector<int> &visit, vector<int> &path, int k, int p_o)
{
for (int i = 0; i < g.size(); i++)
{
if (g[k][i] == 0)
continue;
if (i == p_o)
{
res.push_back(path); //保存找到的環
continue;
}
if (visit[i] == 1)
continue;
visit[i] = 1;
path.push_back(i);
dfs(g, res, visit, path, i, p_o);
path.pop_back();
visit[i] = 0;
}
}
bool cmp(vector<int> a, vector<int> b)
{
//輸出排序比較
if (a.size() == b.size())
{
for (int i = 0; i < a.size(); i++)
{
if (a[i] == b[i])
continue;
return a[i] < b[i];
}
}
else
return a.size() < b.size();
return false;
}
bool isSame(vector<int> &a, vector<int> &b)
{
//比較兩個矩陣是否一樣
if (a.size() != b.size())
return false;
for (int i = 0; i < a.size(); i++)
{
if (a[i] != b[i])
return false;
}
return true;
}
void rot_vector(vector<int> &nums)
{
//統一環的起點,最小id爲起點
int min_num = nums[0];
int min_idx = 0;
for (int i = 0; i < nums.size(); i++)
{
if (min_num > nums[i])
{
min_num = nums[i];
min_idx = i;
}
}
vector<int> temp(nums);
for (int i = 0; i < nums.size(); i++)
{
nums[i] = temp[(i + min_idx) % nums.size()];
}
}
int main()
{
clock_t start_time, end_time;
start_time = clock();
//--------------------------數據讀入-----------------------------
ifstream infile("test_data.txt");
vector<int> ids1;
vector<int> ids2;
vector<int> vals;
int id1, id2, val;
char c;
while (infile >> id1 >> c >> id2 >> c >> val)
{
ids1.push_back(id1);
ids2.push_back(id2);
vals.push_back(val);
}
//-----------------------創建有向圖-----------------------------------
int max_id = 0;
for (int i = 0; i < ids1.size(); i++)
{
if (i == 0 || max_id < ids1[i])
{
max_id = ids1[i];
}
if (max_id < ids2[i])
{
max_id = ids2[i];
}
}
cout << max_id << endl;
max_id += 1;
vector<int> temp(max_id, 0);
vector<vector<int>> g(max_id, temp);
vector<int> visit(max_id, -1);
for (int i = 0; i < ids1.size(); i++)
{
visit[ids1[i]] = 0;
visit[ids2[i]] = 0;
}
for (int i = 0; i < ids1.size(); i++)
{
g[ids1[i]][ids2[i]] = vals[i];
}
//-------------------深度遍歷找環------------------------------------
vector<vector<int>> res;
vector<int> path;
for (int i = 0; i < visit.size(); i++)
{
if (visit[i] == -1)
continue;
visit[i] = 1;
path.push_back(i);
dfs(g, res, visit, path, i, i);
visit[i] = 0;
path.pop_back();
}
//-----------------將環排序去重----------------------------------------
cout << res.size() << endl;
vector<vector<int>> res1;
vector<vector<int>> res2;
for (int i = 0; i < res.size(); i++)
{
if (res[i].size() < 3 || res[i].size() > 7)
continue;
rot_vector(res[i]);
res1.push_back(res[i]);
}
sort(res1.begin(), res1.end(), cmp);
res2.push_back(res1[0]);
for (int i = 1; i < res1.size(); i++)
{
if (isSame(res1[i], res1[i - 1]))
continue;
res2.push_back(res1[i]);
}
//------------------輸出結果--------------------------------------------
ofstream outfile("result.txt");
outfile << res2.size() << endl;
for (int i = 0; i < res2.size(); i++)
{
for (int j = 0; j < res2[i].size(); j++)
outfile << res2[i][j] << ",";
outfile << endl;
}
end_time = clock();
cout << " time : " << double(end_time - start_time) / CLOCKS_PER_SEC << "s" << endl;
}