HDU 3836 Equivalent Sets

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=3836

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 3469    Accepted Submission(s): 1199

Problem Description

To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

 

Input

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

 

Output

For each case, output a single integer: the minimum steps needed.

 

Sample Input

4 0 3 2 1 2 1 3

 

Sample Output

4 2

Hint

Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.

 题意

给你一个有向图，问你至少需要添加多少条边构成强联通。

题解

先Tarjan缩点，然后统计有出度为0和入度为0的点的个数，取个最大值即可，至于为什么，请自行脑补。

代码

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<stack>
#define MAX_N 20002
using namespace std;

vector<int> G[MAX_N];
int dfn[MAX_N],low[MAX_N],ind=0;
bool vis[MAX_N];
bool inStack[MAX_N];

stack<int> st;

int color[MAX_N],tot=0;

vector<int> newG[MAX_N];
vector<int> newrG[MAX_N];

void Tarjan(int u) {
    dfn[u] = low[u] = ++ind;
    vis[u] = 1;
    inStack[u] = 1;
    st.push(u);

    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (!vis[v]) {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (inStack[v])
            low[u] = min(dfn[v], low[u]);
    }
    if (dfn[u] == low[u]) {
        int x;
        do {
            x = st.top();
            st.pop();
            inStack[x] = 0;
            color[x] = tot;
        } while (x != u);
        tot++;
    }
}

int main() {
    int n, m;
    cin.sync_with_stdio(false);
    while (cin >> n >> m) {
        memset(vis, 0, sizeof(vis));
        memset(inStack, 0, sizeof(inStack));
        while (st.size())st.pop();
        for (int i = 0; i <= n; i++)G[i].clear();
        for (int i = 0; i <= n; i++)newG[i].clear();
        for (int i = 0; i <= n; i++)newrG[i].clear();
        ind = 0;
        tot = 0;
        memset(color, 0, sizeof(color));

        for (int i = 0; i < m; i++) {
            int u, v;
            cin >> u >> v;
            G[u].push_back(v);
        }
        for (int i = 1; i <= n; i++)
            if (!vis[i])Tarjan(i);
        for (int i = 1; i <= n; i++)
            for (int j = 0; j < G[i].size(); j++) {
                int u = color[i], v = color[G[i][j]];
                if (u == v)continue;
                newG[u].push_back(v);
                newrG[v].push_back(u);
            }
        if (tot == 1) {
            cout << 0 << endl;
            continue;
        }
        int a = 0, b = 0;
        for (int i = 0; i < tot; i++)if (newG[i].size() == 0)a++;
        for (int i = 0; i < tot; i++)if (newrG[i].size() == 0)b++;
        cout << max(a, b) << endl;
    }
    return 0;
}