JZOJ 4253.QYQ在艾泽拉斯

\(\text{Problem}\)

有向不联通图,求每个子图至多选出一条最大权值和的路径,求前 \(k+1\)

\(\text{Solution}\)

显然将每个子图缩点后 \(dp\),排序 \(dp\) 值即可
很多细节要处理

\(\text{Code}\)

#include<cstdio>
#include<algorithm>
#define re register
using namespace std;

const int N = 1e5 + 5, M = 1e6 + 5; 
int n, m, k, a[N], Q[N], len;

struct edge{
	int to, nxt;
}e[M], E[M], _e[M];
int h[N], H[N], _h[N];

inline void add(int x, int y)
{
	static int tot = 0;
	e[++tot] = edge{y, h[x]}, h[x] = tot;
}
inline void Add(int x, int y)
{
	static int Tot = 0;
	E[++Tot] = edge{y, H[x]}, H[x] = Tot;
}
inline void _add(int x, int y)
{
	static int _tot = 0;
	_e[++_tot] = edge{y, _h[x]}, _h[x] = _tot;
}

int bz[N];
void find(int x)
{
	Q[++len] = x, bz[x] = 1;
	for(re int i = H[x]; i; i = E[i].nxt)
	{
		int v = E[i].to;
		if (bz[v]) continue;
	 	find(v);
	}
}

int dfn[N], low[N], vis[N], st[N], top, dfc, col[N], color;
void tarjan(int x)
{
	st[++top] = x, dfn[x] = low[x] = ++dfc, vis[x] = 1;
	for(re int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (!dfn[v]) tarjan(v), low[x] = min(low[x], low[v]);
		else if (vis[v]) low[x] = min(low[x], dfn[v]);
	}
	if (low[x] == dfn[x])
	{
		col[x] = ++color, vis[x] = 0;
		while (st[top] != x) col[st[top]] = color, vis[st[top]] = 0, --top;
		--top;
	}
}

int _Q[N], f[N], g[N], cnt, in[N], _in[N], cur, val[N];
inline void topu() 
{
	for(re int i = 1; i <= len; i++)
	{
		val[col[Q[i]]] += a[Q[i]];
		for(re int j = h[Q[i]]; j; j = e[j].nxt)
			if (col[e[j].to] != col[Q[i]]) _add(col[Q[i]], col[e[j].to]);
	} 
	for(re int i = cur + 1; i <= color; i++)
		for(re int j = _h[i]; j; j = _e[j].nxt) _in[_e[j].to]++;
		
	int head = 0, tail = 0;
	for(re int i = cur + 1; i <= color; i++)
	{
		f[i] = val[i];
		if (!_in[i]) _Q[++tail] = i;
	}
	while (head < tail)
	{
		int now = _Q[++head];
		for(re int i = _h[now]; i; i = _e[i].nxt)
		{
			f[_e[i].to] = max(f[_e[i].to], f[now] + val[_e[i].to]), --_in[_e[i].to];
			if (!_in[_e[i].to]) _Q[++tail] = _e[i].to;
		}
	}
	++cnt;
	for(re int i = cur + 1; i <= color; i++) g[cnt] = max(g[cnt], f[i]);
	cur = color;
}

int main()
{
	freopen("azeroth.in", "r", stdin);
	freopen("azeroth.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for(re int i = 1, u, v; i <= m; i++)
	{
		scanf("%d%d", &u, &v);
		if (u ^ v) add(u, v), Add(u, v), Add(v, u);
	} 
	for(re int i = 1; i <= n; i++) scanf("%d", &a[i]);
	scanf("%d", &k);
	for(re int i = 1; i <= n; i++)
	if (!bz[i])
	{
		len = top = 0, find(i);
		for(re int j = 1; j <= len; j++)
		if (!dfn[Q[j]]) tarjan(Q[j]);
		topu();
	}
	sort(g + 1, g + cnt + 1), ++k; 
	int ans = 0;
	for(re int i = cnt; i && k; --i, --k) ans += g[i];
	printf("%d\n", ans);
}
posted @ 2021-08-10 21:56  leiyuanze  阅读(23)  评论(0编辑  收藏  举报