1139 First Contact（30 分）
Unlike in nowadays, the way that boys and girls expressing their feelings of love was quite subtle in the early years. When a boy A had a crush on a girl B, he would usually not contact her directly in the first place. Instead, he might ask another boy C, one of his close friends, to ask another girl D, who was a friend of both B and C, to send a message to B -- quite a long shot, isn‘t it? Girls would do analogously.
Here given a network of friendship relations, you are supposed to help a boy or a girl to list all their friends who can possibly help them making the first contact.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (1 < N ≤ 300) and M, being the total number of people and the number of friendship relations, respectively. Then M lines follow, each gives a pair of friends. Here a person is represented by a 4-digit ID. To tell their genders, we use a negative sign to represent girls.
After the relations, a positive integer K (≤ 100) is given, which is the number of queries. Then K lines of queries follow, each gives a pair of lovers, separated by a space. It is assumed that the first one is having a crush on the second one.
Output Specification:
For each query, first print in a line the number of different pairs of friends they can find to help them, then in each line print the IDs of a pair of friends.
If the lovers A and B are of opposite genders, you must first print the friend of A who is of the same gender of A, then the friend of B, who is of the same gender of B. If they are of the same gender, then both friends must be in the same gender as theirs. It is guaranteed that each person has only one gender.
The friends must be printed in non-decreasing order of the first IDs, and for the same first ones, in increasing order of the seconds ones.
Sample Input:
10 18
-2001 1001
-2002 -2001
1004 1001
-2004 -2001
-2003 1005
1005 -2001
1001 -2003
1002 1001
1002 -2004
-2004 1001
1003 -2002
-2003 1003
1004 -2002
-2001 -2003
1001 1003
1003 -2001
1002 -2001
-2002 -2003
5
1001 -2001
-2003 1001
1005 -2001
-2002 -2004
1111 -2003
Sample Output:
4 1002 2004 1003 2002 1003 2003 1004 2002 4 2001 1002 2001 1003 2002 1003 2002 1004 0 1 2003 2001 0
當初考試的時候只得了一部分的分，現在拿出來重新做，思路清晰了很多，對於A，找一個同性好友C，C再找個好友D，D要是B的同性好友，男女編號絕對值可能一樣，所以女編號用加10000來區分，但是0000和-0000，通過int讀取是無法區分性別的，所以選擇字符串讀取。
用vector存某人的好友，也方便查找給定一個編號判斷是否有這個好友。
代碼：
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; typedef pair<int,int> P; vector<int> s[20000];///存好友 下標代表本人 P ans[90000];///存答案 int vis[20000];///標記是否排序過 int n,m,k,num; char a[6],b[6]; inline int change(char* t) {///字符串轉換為int int d = 0,i = 0; if(t[0] == ‘-‘)d = 1,i ++;///負數直接絕對值加10000來區分 while(t[i]) { d = d * 10 + t[i ++] - ‘0‘; } return d; } inline bool check(int x,int y) {///檢查是否同性 以10000為界限 if(x >= 10000 && y >= 10000 || x < 10000 && y < 10000)return true; return false; } int main() { scanf("%d%d",&n,&m); for(int i = 0;i < m;i ++) { scanf("%s%s",a,b); int aa = change(a); int bb = change(b); s[aa].push_back(bb); s[bb].push_back(aa); } scanf("%d",&k); while(k --) { scanf("%s%s",a,b); int aa = change(a); int bb = change(b); if(!vis[aa]) { vis[aa] = 1; sort(s[aa].begin(),s[aa].end()); } num = 0; for(int i = 0;i < s[aa].size();i ++) { if(s[aa][i] != bb && check(aa,s[aa][i])) {///先找A的同性好友C 要保證不是B AB可能同性 int t = s[aa][i]; if(!vis[t]) { vis[t] = 1; sort(s[t].begin(),s[t].end()); } for(int j = 0;j < s[t].size();j ++) { ///找C的好友D 保證D不等於A 如果D和B同性 且是B好友則滿足 if(s[t][j] != aa && check(bb,s[t][j]) && find(s[s[t][j]].begin(),s[s[t][j]].end(),bb) != s[s[t][j]].end()) { ans[num ++] = P(t,s[t][j]); } } } } printf("%d\n",num); for(int i = 0;i < num;i ++) { printf("%04d %04d\n",ans[i].first % 10000,ans[i].second % 10000); } } }
1139 First Contact（30 分）