洛谷1345 [USACO5.4]奶牛的电信Telecowmunication

原题链接

最小割点数转换成最小割边数的模板题（不过这数据好小）。
每个点拆成两个点，连一条容量为(1)的边，原图的边容量定为(+infty)，然后跑最大流即可。
这里用的是(Dinic)。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 220;
const int M = 5e3;
int fi[N], di[M], ne[M], da[M], de[N], q[N], cu[N], l = 1, st, ed;
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline void add(int x, int y, int z)
{
	di[++l] = y;
	da[l] = z;
	ne[l] = fi[x];
	fi[x] = l;
	di[++l] = x;
	da[l] = 0;
	ne[l] = fi[y];
	fi[y] = l;
}
inline int minn(int x, int y){ return x < y ? x : y; }
bool bfs()
{
	int i, x, y, head = 0, tail = 1;
	memset(de, 0, sizeof(de));
	q[1] = st;
	de[st] = 1;
	while (head ^ tail)
	{
		x = q[++head];
		for (i = fi[x]; i; i = ne[i])
			if (!de[y = di[i]] && da[i] > 0)
			{
				de[y] = de[x] + 1;
				if (!(y ^ ed))
					return true;
				q[++tail] = y;
			}
	}
	return false;
}
int dfs(int x, int k)
{
	if (!(x ^ ed))
		return k;
	int mi, y;
	for (int &i = cu[x]; i; i = ne[i])
		if (!(de[y = di[i]] ^ (de[x] + 1)) && da[i] > 0)
		{
			mi = dfs(y, minn(k, da[i]));
			if (mi > 0)
			{
				da[i] -= mi;
				da[i ^ 1] += mi;
				return mi;
			}
		}
	return 0;
}
int main()
{
	int i, n, m, x, y, s = 0;
	n = re();
	m = re();
	st = re() + n;
	ed = re();
	for (i = 1; i <= n; i++)
		add(i, i + n, 1);
	for (i = 1; i <= m; i++)
	{
		x = re();
		y = re();
		add(x + n, y, 1e9);
		add(y + n, x, 1e9);
	}
	while (bfs())
	{
		for (i = 1; i <= (n << 1); i++)
			cu[i] = fi[i];
		for (; (x = dfs(st, 1e9)) > 0; s += x);
	}
	printf("%d", s);
	return 0;
}