#include<bits/stdc++.h>
using namespace std;
vector<int> graph[10005];
bool boy[10005];            // !!!妙啊 下标为id true男生 false女生
int N, M, K;
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    scanf("%d%d", &N, &M);
    string s1, s2;
    int A, B;
    fill(boy, boy+10005, false);
    for(int i = 0; i < M; i++){
        cin>>s1>>s2;
        A = abs(stoi(s1));B = abs(stoi(s2));            // !!!绝对值
        boy[A] = (s1[0] != '-');
        boy[B] = (s2[0] != '-');
        graph[A].push_back(B);
        graph[B].push_back(A);
    }
    scanf("%d", &K);
    vector<pair<int, int>>ans;
    while(K--){
        scanf("%d%d", &A, &B);
        A = abs(A);B = abs(B);
        ans.clear();
        for(auto C : graph[A]){
            if(C != A && C != B && boy[C] == boy[A]){           // !!!C != A && C != B
                for(auto D : graph[C]){
                    if(D != A && D != B && boy[D] == boy[B]){   // !!!D != A && D != B
                        if(find(graph[D].begin(), graph[D].end(), B) != graph[D].end())
                            ans.push_back({C, D});              // !!!注意加大括号
                    }
                }
            }
        }
        printf("%d\n", ans.size());
        sort(ans.begin(), ans.end());                   // !!!妙啊
        for(auto it : ans)
            printf("%04d %04d\n", it.first, it.second);
    }
    return 0;
}

4，解题过程

第一搏

没想太多，unordered_map<int, unordered_set<int>> relation存储每个人的朋友关系，两个for循环加一个find函数。。。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int a, b;
};
unordered_map<int, unordered_set<int>> relation;
vector<node> ans;
bool cmp1(node x, node y){
    return abs(x.a) != abs(y.a) ? abs(x.a) < abs(y.a) : abs(x.b) < abs(y.b);
}
void getAns(int a, int b){
    unordered_set<int> temp1 = relation[a];
    for(auto c : temp1){
        if(a * c > 0){//a->c
            unordered_set<int> temp2 = relation[c];
            for(auto d : temp2){//c->d
                if(d * b > 0 &&
                   find(relation[d].begin(), relation[d].end(), b) != relation[d].end())//d->b
                    ans.push_back({c, d});
            }
        }
    }
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    int N, M, K, a, b;//N节点数目 M朋友关系数目 K查询数目
    scanf("%d%d", &N, &M);
    for(int i = 0; i < M; i++){
        scanf("%d%d", &a, &b);
        relation[a].insert(b);
        relation[b].insert(a);
    }
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%d", &a, &b);
        ans.clear();
        getAns(a, b);
        sort(ans.begin(), ans.end(), cmp1);
        printf("%d\n", ans.size());
        for(int j = 0; j < ans.size(); j++)
            printf("%04d %04d\n", abs(ans[j].a), abs(ans[j].b));
        //if(ans.size() != 0) printf("\n");
    }
    return 0;
}

*PAT_甲级_1139 First Contact (30point(s)) (C++)【图的遍历】_甲级_04

第二搏

忽略了-0000这种情况。。。

认真看了看代码，感觉改起来太难了

但还是决定试一试

flag记录0000/-0000的符号，flagA/flagB/flagC/flagD记录a/b/c/d的符号。。。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int a, b;
};
unordered_map<int, unordered_set<int>> relation;
vector<node> ans;
int flag = 0;//1男生 -1女生
bool cmp1(node x, node y){
    return abs(x.a) != abs(y.a) ? abs(x.a) < abs(y.a) : abs(x.b) < abs(y.b);
}
void getAns(int a, int b){
    int flagA, flagB, flagC, flagD;
    if(a == 0) flagA = flag;
    else flagA = a>0 ? 1:-1;
    if(b == 0) flagB = flag;
    else flagB = b>0 ? 1:-1;
    for(auto c : relation[a]){
        if(c == 0) flagC = flag;
        else flagC = c>0 ? 1:-1;
        if(flagA * flagC > 0){//a->c
            for(auto d : relation[c]){//c->d
                if(d == 0) flagD = flag;
                else flagD = d>0 ? 1:-1;
                if(flagB * flagD > 0 &&
                   find(relation[d].begin(), relation[d].end(), b) != relation[d].end())//d->b
                    ans.push_back({c, d});
            }
        }
    }
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    int N, M, K, a, b;//N节点数目 M朋友关系数目 K查询数目
    string s1, s2;
    scanf("%d%d", &N, &M);
    for(int i = 0; i < M; i++){
        cin>>s1>>s2;
        a = stoi(s1);b = stoi(s2);
        if(a == 0 && s1[0] == '-') flag = -1;
        else flag = 1;
        if(b == 0 && s2[0] == '-') flag = -1;
        else flag = 1;
        relation[a].insert(b);
        relation[b].insert(a);
    }
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%d", &a, &b);

        ans.clear();
        getAns(a, b);
        sort(ans.begin(), ans.end(), cmp1);
        printf("%d\n", ans.size());
        for(int j = 0; j < ans.size(); j++)
            printf("%04d %04d\n", abs(ans[j].a), abs(ans[j].b));
        //if(ans.size() != 0) printf("\n");
    }
    return 0;
}