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 = 0; i < adj[here].size(); ++i) {
        int there = adj[here][i];
        //(here,there)是树边
        if (discovered[there] == -1)
            ret = min(ret, scc(there));
        else if (discovered[there] < discovered[here] && finished[there] != 1)
            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] = 1;
    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(), -1);
    sccCounter = vertexCounter = 0;

    //对所有顶点调用scc()
    for (int i = 0; i < adj.size(); ++i)
        if (discovered[i] == -1) 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 * 2);
    for (int i = 0; i < vars; i += 2) {
        //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 + 1;
        adj[i * 2 + 1].push_back(j * 2);
        adj[j * 2 + 1].push_back(i * 2);
    }

    for (int i = 0; i < vars; ++i) {
        for (int j = 0; 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 * 2].push_back(j * 2 + 1);
                adj[j * 2].push_back(i * 2 + 1);
            }
        }
    }
}


vector<int> solve2SAT()
{
    int n = adj.size() / 2;
    vector<int> label = tarjanSCC();

    for (int i = 0; i < 2 * n; i += 2)
        if (label[i] == label[i + 1])
            return vector<int>();

    vector<int> value(2 * n, -1);

    vector<pair<int, int>> order;
    for (int i = 0; i < 2 * n; i++)
        order.push_back(make_pair(label[i], i));
    sort(order.begin(), order.end());

    for (int i = 0; i < 2 * n; ++i) {
        int vertex = order[i].second;
        int variable = vertex / 2, isTrue = vertex % 2;
        if (value[variable] != -1) 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 = 0; i < v.size()/2; i+=2) {
                if (v[i] == 1) {
                    cout << meetings[i].first << ' ' << meetings[i].second << endl;
                }
                else {
                    cout << meetings[i+1].first << ' ' << meetings[i+1].second << endl;
                }
            }
        }
    }


    return 0;
}

 

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 + 7;
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] = 1;

    for (int i = 0; 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 = 0;
        while (1) {
            int now = s.top();
            s.pop();
            vis[x] = 0;
            cnt++;
            if (now == x) break;
        }
        if (cnt > 1) ans = min(ans, cnt);
    }

}

int main()
{
    scanf("%d",&n );
    for (int i = 1; i <= n; i++) {
        int x;
        scanf("%d", &x);
        E[i].push_back(x);
    }

    for (int i = 1; i <= n; i++) {
        if (!dfn[i])
            tarjan(i);
    }

    cout << ans << endl;

    return 0;
}