POJ1236DAG转强连通图

Posted 2021-08-20 hesorchen

题目

POJ 1236

不难得出第一问答案是，对原图tarjan缩点、转化为DAG之后入度为0的点个数。

第二问有个结论：要将DAG图转化为强连通图，需要增加的最少单向边数量为max(indegree,outdegree)。

因为强连通图中每个点对都要互相可达，那就每个点都要可进可出。

代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <map>
using namespace std;

const int N = 105;
vector<int> edge[N];
int dfn[N], low[N], ct, cnt;
bool instack[N];
vector<int> st;
vector<int> blocks[N];
map<int, int> mp;
void tarjan(int u)
{
    dfn[u] = low[u] = ++ct;
    instack[u] = 1;
    st.push_back(u);
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u])
    {
        ++cnt;
        int node;
        do
        {
            node = st.back();
            st.pop_back();
            instack[node] = 0;
            mp[node] = cnt;
            blocks[cnt].push_back(node);
        } while (low[node] != dfn[node]);
    }
}

int indegree[N];
int outdegree[N];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int v;
        while (cin >> v, v)
            edge[i].push_back(v);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j < edge[i].size(); j++)
        {
            int it = edge[i][j];
            if (mp[i] == mp[it])
                continue;
            indegree[mp[it]] = 1;
            outdegree[mp[i]] = 1;
        }
    }
    int res1 = 0, res2 = 0;
    for (int i = 1; i <= n; i++)
        if (blocks[i].size())
            (res1 += !outdegree[i]), (res2 += !indegree[i]);
    cout << res2 << endl
         << (cnt == 1 ? 0 : max(res1, res2)) << endl;
    return 0;
}