CH#17C 舞动的夜晚

原题链接

即求二分图的不可行边数量，因为不保证是完备匹配，所以需要通过网络流求出任意一组最大匹配，并建立新图判断。
建新图：对于跑完网络流的图上已经匹配的边，建立反边；对于没有匹配的边，建立正边（图只改变边的方向，别的结构不变）。
有结论：

1. 必须边的判定条件为：\((x,y)\)的流量为\(1\)，并且在残量网络上属于不同的强联通分量。
2. 可行边的判定条件为：\((x,y)\)的流量为\(1\)，或者在残量网络上属于同一个强联通分量。

所以我们在新图上跑\(tarjan\)，再逐边检验即可。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 2e4 + 10;
const int M = 2e5 + 10;
struct dd {
    int x, y;
};
dd a[M];
int fi[N], di[M << 1], ne[M << 1], da[M << 1], cfi[N], cdi[M], cne[M], dfn[N], low[N], sta[N], bl[N], cu[N], de[N], q[M << 1], an[M], l = 1, tp, lc, SCC, st, ed, ti;
bool v[N];
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;
}
inline void add_c(int x, int y)
{
    cdi[++lc] = y;
    cne[lc] = cfi[x];
    cfi[x] = lc;
}
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)
{
    int y, mi;
    if (!(x ^ ed))
        return k;
    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;
}
void tarjan(int x)
{
    int i, y;
    dfn[x] = low[x] = ++ti;
    sta[++tp] = x;
    v[x] = 1;
    for (i = cfi[x]; i; i = cne[i])
    {
        if (!dfn[y = cdi[i]])
        {
            tarjan(y);
            low[x] = minn(low[x], low[y]);
        }
        else
            if (v[y])
                low[x] = minn(low[x], dfn[y]);
    }
    if (!(dfn[x] ^ low[x]))
    {
        SCC++;
        do
        {
            y = sta[tp--];
            bl[y] = SCC;
            v[y] = 0;
        } while (x ^ y);
    }
}
int main()
{
    int i, n, m, k, x, y, s = 0;
    n = re();
    m = re();
    k = re();
    st = n + m + 1;
    ed = st + 1;
    for (i = 1; i <= k; i++)
    {
        x = re();
        y = re() + n;
        a[i].x = x;
        a[i].y = y;
        add(x, y, 1);
        add(y, x, 0);
    }
    for (i = 1; i <= n; i++)
    {
        add(st, i, 1);
        add(i, st, 0);
    }
    for (i = 1; i <= m; i++)
    {
        add(i + n, ed, 1);
        add(ed, i + n, 0);
    }
    while (bfs())
    {
        for (i = 1; i <= ed; i++)
            cu[i] = fi[i];
        while (dfs(st, 1e9) > 0);
    }
    for (i = 2; i <= l; i++)
        if (da[i])
            add_c(di[i ^ 1], di[i]);
    for (i = 1; i <= ed; i++)
        if (!dfn[i])
            tarjan(i);
    for (i = 1; i <= k;i++)
        if (bl[a[i].x] ^ bl[a[i].y] && da[i << 1])
            an[++s] = i;
    printf("%d\n", s);
    if (!s)
    {
        printf("\n");
        return 0;
    }
    for (i = 1; i < s; i++)
        printf("%d ", an[i]);
    printf("%d", an[s]);
    return 0;
}