pat1139 First Contact

题意：a要和b和好，首先a要找同性的c，c找d，d和b同性，这样就能和好，输出所有的可能。

思路：找到所有和a同性的人，再找到所有和b同性的人，然后看这两个集合中有哪两个人有联系，当然ab不能直接联系。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>

using namespace std;

const int MAX_N = 310;
vector<string> v;
map<string, int> mp;

int get_id(string x) {
    if (mp.count(x)) return mp[x];
    v.push_back(x);
    return mp[x] = v.size()-1;
}

int N, M, Q;
int gen[MAX_N];
char s1[100], s2[100];
vector<int> e[MAX_N][2];
typedef pair<string, string> P;
typedef pair<int, int> pi;
vector<P> ans;
map<pi, bool> mt;

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%d %d", &N, &M);
    for (int i = 0; i < M; i++) {
        scanf("%s %s", s1, s2);
        if (s1[0]=='-') {gen[get_id(s1)]=1;}
        if (s2[0]=='-') {gen[get_id(s2)]=1;}
        int id1 = get_id(s1), id2 = get_id(s2);
        e[id1][gen[id1]!=gen[id2]].push_back(id2);
        e[id2][gen[id1]!=gen[id2]].push_back(id1);
        mt[make_pair(id1, id2)] = 1;
        mt[make_pair(id2, id1)] = 1;
    }
    scanf("%d", &Q);
    while (Q--) {
        ans.clear();
        scanf("%s %s", s1, s2);
        int id1 = get_id(s1), id2 = get_id(s2);
        vector<int> v1, v2;
        for (int i = 0; i < e[id2][0].size(); i++) {
            if (e[id2][0][i] == id1) continue;
            v2.push_back(e[id2][0][i]);
        }
        for (int i = 0; i < e[id1][0].size(); i++) {
            if (e[id1][0][i] == id2) continue;
            v1.push_back(e[id1][0][i]);
        }
        for (int i = 0; i < v1.size(); i++) {
            for (int j = 0; j < v2.size(); j++) {
                if (mt[make_pair(v1[i], v2[j])]) {
                    string ss = v[v1[i]], sq = v[v2[j]];
                    if (ss[0]=='-') ss=ss.substr(1);
                    if (sq[0]=='-') sq=sq.substr(1);
                    ans.push_back(make_pair(ss, sq));
                }
            }
        }
        sort(ans.begin(), ans.end());
        printf("%lu\n", ans.size());
        for (int i = 0; i < ans.size(); i++) {
            string ss = ans[i].first, sq = ans[i].second;
            printf("%s %s\n", ss.c_str(), sq.c_str());
        }
    }
    return 0;
}