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;
}