算法问题实战策略 MEETINGROOM 附一份tarjan模板

地址 https://algospot.com/judge/problem/read/MEETINGROOM

解答  2-sat 代码样例过了 没有ac。 我又没有正确代码对拍。。。。。

已确认是输出问题 修改完成

 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <stack>

 using namespace std;

 vector<vector<int>> adj;

 vector<int> sccId, discovered, finished;
 stack<int> st;    //保存顶点序号的栈
 int sccCounter, vertexCounter;

 //返回以here为根节点的子树中
 //能够到达后向边的最小发现顺序
 int scc(int here) {
     int ret = discovered[here] = vertexCounter++;
     //将here存入栈，here的所有后代节点都会在here之后进栈
     st.push(here);

     for (int i = ; i < adj[here].size(); ++i) {
         int there = adj[here][i];
         //(here,there)是树边
         if (discovered[there] == -)
             ret = min(ret, scc(there));
         else if (discovered[there] < discovered[here] && finished[there] != )
             ret = min(ret, discovered[there]);
     }

     //判断here是否为强联通分量的根节点
     if (ret == discovered[here]) {
         //以here为根节点的子树中，将剩余所有顶点全部绑定为同一分量
         while (true) {
             int t = st.top();
             st.pop();
             sccId[t] = sccCounter;
             if (t == here) break;
         }
         ++sccCounter;
     }

     finished[here] = ;
     return ret;
 }

 //tarjan 的scc算法
 vector<int> tarjanSCC() {
     //数组和计数器的初始化
     sccId = discovered = finished = vector<int>(adj.size(), -);
     sccCounter = vertexCounter = ;

     //对所有顶点调用scc()
     for (int i = ; i < adj.size(); ++i)
         if (discovered[i] == -) scc(i);
     return sccId;
 }

 //========================================================================
 //图的领接表表示法
 //vector<vector<int>> adj;

 bool disjoint(const pair<int, int>& a, const pair<int, int>& b) {
     return a.second <= b.first || b.second <= a.first;
 }

 //如果meetings[]表示各队提出的开会时间
 //则将此题转换为2-SAT问题后生成蕴含图
 //第i个团队需要选择meetings[2*i]或meetings[2*i+1]时候之一开会
 void makeGraph(const vector<pair<int, int>>& meetings)
 {
     int vars = meetings.size();

     //每个变量对应图的两个顶点
     adj.clear(); adj.resize(vars * );
     for (int i = ; i < vars; i += ) {
         //各团队需要选择第i号和第j号会议之一
         //添加(i or j )句子
         int j = i + ;
         adj[i *  + ].push_back(j * );
         adj[j *  + ].push_back(i * );
     }

     for (int i = ; i < vars; ++i) {
         for (int j = ; j < i; ++j) {
             //第i号会议和第j号会议重叠
             if (!disjoint(meetings[i], meetings[j])) {
                 //放弃第i个会议 或者放弃第j个会议
                 //添加 (~i or ~j)子句
                 adj[i * ].push_back(j *  + );
                 adj[j * ].push_back(i *  + );
             }
         }
     }
 }

 vector<int> solve2SAT()
 {
     int n = adj.size() / ;
     vector<int> label = tarjanSCC();

     for (int i = ; i <  * n; i += )
         if (label[i] == label[i + ])
             return vector<int>();

     vector<int> value( * n, -);

     vector<pair<int, int>> order;
     for (int i = ; i <  * n; i++)
         order.push_back(make_pair(label[i], i));
     sort(order.begin(), order.end());

     for (int i = ; i <  * n; ++i) {
         int vertex = order[i].second;
         int variable = vertex / , isTrue = vertex % ;
         if (value[variable] != -) continue;
         value[variable] = !isTrue;
     }
     return value;
 }

 int main()
 {
     int n;
     cin >> n;

     while (n--) {
         int m;
         cin >> m;
         vector<pair<int, int>> meetings;
         while (m--) {
             int a, b, c, d;
             cin >> a >> b >> c >> d;
             meetings.push_back(make_pair(a, b));
             meetings.push_back(make_pair(c, d));
         }
         makeGraph(meetings);

         vector<int> v = solve2SAT();
         if (v.empty()) {
             cout << "IMPOSSIBLE" << endl;;
         }
         else {
             cout << "POSSIBLE" << endl;
             for (int i = ; i < v.size()/; i+=) {
                 if (v[i] == ) {
                     cout << meetings[i].first << ' ' << meetings[i].second << endl;
                 }
                 else {
                     cout << meetings[i+].first << ' ' << meetings[i+].second << endl;
                 }
             }
         }
     }

     return ;
 }

一份tarjan 模板

 // 11111111.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
 //

 #include "pch.h"
 #include <iostream>
 #include <vector>
 #include <stack>
 #include <algorithm>

 using namespace std;

 const int maxn = 2e5 + ;
 vector<int> E[maxn];
 int vis[maxn];
 int dfn[maxn], low[maxn], tot, n, ans = maxn;
 stack<int> s;

 void tarjan(int x)
 {
     low[x] = dfn[x] = ++ tot;
     s.push(x); vis[x] = ;

     for (int i = ; i < E[x].size(); i++) {
         int v = E[x][i];
         if (!dfn[v]) {
             tarjan(v);
             low[x] = min(low[x], low[v]);
         }
         else if (vis[v]) {
             low[x] = min(low[x], dfn[v]);
         }
     }

     if (low[x] == dfn[x]) {
         int cnt = ;
         while () {
             int now = s.top();
             s.pop();
             vis[x] = ;
             cnt++;
             if (now == x) break;
         }
         if (cnt > ) ans = min(ans, cnt);
     }

 }

 int main()
 {
     scanf("%d",&n );
     for (int i = ; i <= n; i++) {
         int x;
         scanf("%d", &x);
         E[i].push_back(x);
     }

     for (int i = ; i <= n; i++) {
         if (!dfn[i])
             tarjan(i);
     }

     cout << ans << endl;

     return ;
 }

tarjan模板