題目大意
假設A是男生,B是女生。A看上了B,於是A先去找同性基友C,C找靚妹D,D轉告B。要保證C和A是同性,D和B是同性。如果A和B是同性,則C和D也和他們同性。
思路解析
本題有個坑就是要注意-0000的情況。所以不能單純的用int處理,要用string,長度不同的就是異性。所以要仔細審題啊,考試時沒有時間推倒重來……示例代碼
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
using namespace std;
map<int, int> mapp;//記錄路徑
vector<vector<int>> vec(10000, vector<int>()), res;
bool cmp(vector<int> v1, vector<int> v2) {
return v1[0] != v2[0] ? v1[0] < v2[0] : v1[1] < v2[1];
}
int main() {
int n, m, k;
scanf("%d %d", &n, &m);
for (int i = 0; i < m; i++) {
string a, b;
cin >> a >> b;
if (a.length() == b.length()) {//爲了防止-0000的情況,不能用int
vec[abs(stoi(a))].push_back(abs(stoi(b)));
vec[abs(stoi(b))].push_back(abs(stoi(a)));
}
mapp[abs(stoi(a)) * 10000 + abs(stoi(b))] = mapp[abs(stoi(b)) * 10000 + abs(stoi(a))] = 1;
}
scanf("%d", &k);
for (int i = 0; i < k; i++) {
int a, b;
scanf("%d %d", &a, &b);
res.clear();
for (int j = 0; j < vec[abs(a)].size(); j++) {
int c = vec[abs(a)][j];
if (c == abs(b))continue;
for (int k = 0; k < vec[abs(b)].size(); k++) {
int d = vec[abs(b)][k];
if (d == abs(a))continue;
if (mapp[abs(c) * 10000 + abs(d)] == 1) {
vector<int> v = { abs(c),abs(d) };
res.push_back(v);
}
}
}
sort(res.begin(), res.end(), cmp);
printf("%d\n", res.size());
for (int j = 0; j < res.size(); j++) {
printf("%04d %04d\n", res[j][0], res[j][1]);
}
}
return 0;
}