一名苦逼的OIer,想成为ACMer

Iowa_Battleship

BZOJ1093或洛谷2272 [ZJOI2007]最大半连通子图

BZOJ原题链接

洛谷原题链接

和 Going from u to v or from v to u?(题解)这道题类似,只不过是求最大子图的大小和个数而已。
一样用\(tarjan\)求强连通分量,并进行缩点,然后对于缩点后的\(DAG\)进行拓扑排序\(DP\)
定义\(size[i]\)表示缩点后的图中每个点(即强连通分量)包含原有的点数,\(f[i]\)表示最大子图(缩点后实际上是一条链)的大小,\(g[i]\)表示大小为\(f[i]\)的终点为\(i\)的子图个数。
设当前边为\((x,y)\),则有转移方程:

\(\qquad\qquad g[y] = g[y] + g[x]\quad \mathrm{if}\ f[x] + size[y]=f[y]\)

\(\qquad\qquad g[y] = g[x],f[y]=f[x]+size[y]\quad \mathrm{if}\ f[x] + size[y]>f[y]\)

最后对拓扑排序\(DP\)后的\(f,g\)数组进行统计即可。
另外,在缩点时必须去重边,这里我用的是\(Hash\)去重,结果因为冲突调了一晚上。。非酋没办法。。

#include<cstdio>
#include<cstring>
#include<bitset>
using namespace std;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
const int P = 1e8 - 11;
struct eg {
	int x, y;
};
eg a[M];
int fi[N], di[M], ne[M], cfi[N], cdi[M], cne[M], dfn[N], low[N], dis[N], sta[N], bl[N], ru[N], si[N], g[N], f[N], q[M], lc, l, tp, SCC, ti, mod;
bool v[N];
bitset<100000000>bs;
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)
{
	di[++l] = y;
	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;
}
void tarjan(int x)
{
	int i, y;
	dfn[x] = low[x] = ++ti;
	sta[++tp] = x;
	v[x] = 1;
	for (i = fi[x]; i; i = ne[i])
		if (!dfn[y = di[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;
			si[SCC]++;
		} while (x ^ y);
	}
}
void topsort()
{
	int i, y, x, head = 0, tail = 0;
	for (i = 1; i <= SCC; i++)
		if (!ru[i])
		{
			q[++tail] = i;
			f[i] = si[i];
			g[i] = 1;
		}
	while (head ^ tail)
	{
		x = q[++head];
		for (i = cfi[x]; i; i = cne[i])
		{
			y = cdi[i];
			ru[y]--;
			if (!ru[y])
				q[++tail] = y;
			if (!((f[x] + si[y]) ^ f[y]))
				g[y] = (g[y] + g[x]) % mod;
			else
				if (f[x] + si[y] > f[y])
				{
					f[y] = f[x] + si[y];
					g[y] = g[x];
				}
		}
	}
}
int hs(int x, int y)
{
	return (12598LL * x + y) % P + 1;
}
int main()
{
	int i, n, m, x, y, k, S = 0, L;
	n = re();
	m = re();
	mod = re();
	for (i = 1; i <= m; i++)
	{
		a[i].x = re();
		a[i].y = re();
		add(a[i].x, a[i].y);
	}
	for (i = 1; i <= n; i++)
		if (!dfn[i])
			tarjan(i);
	for (i = 1; i <= m; i++)
	{
		x = bl[a[i].x];
		y = bl[a[i].y];
		if (x ^ y && !bs[k = hs(x, y)])
		{
			add_c(x, y);
			bs[k] = 1;
			ru[y]++;
		}
	}
	topsort();
	for (i = 1; i <= SCC; i++)
		if (S < f[i])
		{
			S = f[i];
			L = g[i];
		}
		else
			if (!(S ^ f[i]))
				L = (L + g[i]) % mod;
	printf("%d\n%d", S, L);
	return 0;
}

posted on 2018-09-20 21:16  Iowa_Battleship  阅读(...)  评论(...编辑  收藏

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