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【网络流24题】最小路径覆盖问题

2018-01-15 20:26 387 查看

Description

click me

Solution

性质:DAG最小路径覆盖数等于对应的二分图的最大二分匹配数。

最近写网络流写得想吐的,还是匈牙利清新一些

Code

/**************************
* Au: Hany01
* Date: Jan 15th, 2018
* Prob: luogu2764
* Email: hany01@foxmail.com
**************************/

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File()
{
#ifdef hany01
freopen("luogu2764.in", "r", stdin);
freopen("luogu2764.out", "w", stdout);
#endif
}

const int maxn = 155, maxm = 6005;

int n, m, v[maxm], nex[maxm], beg[maxn], mat[maxn], vis[maxn], e, Ans, ismated[maxn];

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

inline void Init()
{
register int uu, vv;
n = read(); m = read();
while (m --) uu = read(), vv = read(), add(uu, vv);
}

bool dfs(int u)
{
for (register int i = beg[u]; i; i = nex[i]) if (!vis[v[i]]) {
vis[v[i]] = 1;
if (!mat[v[i]] || dfs(mat[v[i]])) {
mat[v[i]] = u;
return 1;
}
}
return 0;
}

inline void Hungary()
{
For(i, 1, n) {
Set(vis, 0);
if (dfs(i)) ++ Ans;
}
}

void findpre(int u)
{
if (mat[u]) findpre(mat[u]);
printf("%d ", u);
}

inline void Print()
{
For(i, 1, n) ismated[mat[i]] = 1;
For(i, 1, n) if (!ismated[i]) findpre(i), putchar('\n');
printf("%d\n", n - Ans);
}

int main()
{
File();
Init();
Hungary();
Print();
return 0;
}
//令公桃李满天下,何用堂前更种花。
//    -- 白居易《奉和令公绿野堂种花》
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