The Maximum Unreachable Node Set 【17南宁区域赛】 【二分匹配】

题目链接

https://nanti.jisuanke.com/t/19979

题意

给出n个点 m 条边 求选出最大的点数使得这个点集之间 任意两点不可达 题目中给的边是有向边

思路

这道题 实际上是求 二分图的最大独立集

二分图的最大独立集 = 顶点数 - 二分图最大匹配

相关概念：
https://blog.csdn.net/whosemario/article/details/8513836

那操作就是

先Flyod 跑出 可达矩阵 再二分匹配

答案就是 n - res

AC代码

#pragma comment(linker, "/STACK:102400000,102400000")

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define pb push_back
#define fi first
#define se second
#define L(on) ((on)<<1)
#define R(on) (L(on) | 1)
#define mkp(a, b) make_pair(a, b)
#define bug puts("***bug***");
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define CLR(a, b) memset(a, (b), sizeof(a));
#define syn_close ios::sync_with_stdio(false); cin.tie(0);
#define sp system("pause");
//#define gets gets_s 

using namespace std;

typedef long long ll;
typedef long double ld;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <double, double> pdd;
typedef pair  pll;
typedef vector <int> vi;
typedef vector  vll;
typedef vector < vi > vvi;

const double PI = acos(-1.0);
const double EI = exp(1.0);
const double eps = 1e-8;

inline int read()
{
    char c = getchar(); int ans = 0, vis = 1;
    while (c < '0' || c > '9') { if (c == '-') vis = -vis;  c = getchar(); }
    while (c >= '0' && c <= '9') { ans = ans * 10 + c - '0'; c = getchar(); }
    return ans * vis;
}

const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fll;
const int maxn = (int)1e2 + 10;
const int MAXN = (int)1e4 + 10;
const ll MOD = (ll)1e9 + 7;

int G[maxn][maxn];
int n, m;

void input()
{
    n = read(), m = read();
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            G[i][j] = 0;
    int x, y;
    for (int i = 0; i < m; i++)
    {
        x = read() - 1, y = read() - 1;
        G[x][y] = 1;
    }
}

void Floyd()
{
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                G[i][j] = (G[i][j] || G[i][k] && G[k][j]);
}

int uN, vN;
int linker[maxn];
bool used[maxn];

bool dfs(int u)
{
    for (int v = 0; v < vN; v++)
        if (G[u][v] && !used[v])
        {
            used[v] = true;
            if (linker[v] == -1 || dfs(linker[v]))
            {
                linker[v] = u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res = 0;
    CLR(linker, -1);
    for (int u = 0; u < uN; u++)
    {
        CLR(used, false);
        if (dfs(u))
            res++;
    }
    return res;
}

void solve()
{
    Floyd(); uN = vN = n;
    printf("%d\n", n - hungary());
}

int main()
{
    int t = read();
    while (t--)
    {
        input(); solve();
    }
}