Tarjan 杂题

P5676 [GZOI2017] 小z玩游戏

将有趣程度作为点来建图相当于是看哪些游戏的 $w [i]$ 和 $e [i]$ 在同一个强连通分量中

这样是 $O (n^{2})$ 的无法通过

考虑换一种建边方式

若当前兴奋值为 $a$ ，某个游戏有趣程度为 $w$ ，兴奋程度为 $e$ ， $w$ 是 $a$ 的倍数。

可以视为兴奋值先从 $a$ 变成了它的倍数 $w$ ，再变成 $e$ 。

那我们就可以对每个数向它的倍数建边，再对于每个游戏从 $w$ 向 $e$ 建边。

跑一遍 $T a r j a n$ ，对于每个游戏判断 $w$ 和 $e$ 是否在同一强连通分量即可。

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 5;
const int maxn = 1e5;
int read()
{
    int x = 0 , f = 1;
    char ch = getchar();
    while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
    while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
    return x * f;
}

int n , ee[N] , w[N] , ans;

int low[N] , dfn[N] , timer , in[N] , id[N] , tot , sta[N] , top;

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

void tarjan ( int u )
{
    dfn[u] = low[u] = ++ timer;
    sta[++top] = u , in[u] = 1;
    for ( auto v : e[u] )
    {
        if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
        else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
    }
    if ( low[u] == dfn[u] )
    {
        ++ tot;
        while ( top )
        {
            int x = sta[top--];
            id[x] = tot , in[x] = 0;
            if ( x == u ) break;
        }
    }
}

void init()
{
    for ( int i = 1 ; i <= maxn ; i ++ ) e[i].clear(); 
    ans = timer = top = tot = 0;
    memset ( dfn , 0 , sizeof dfn );
    memset ( id , 0 , sizeof id );
    memset ( in , 0 , sizeof in );
}



signed main ()
{
    ios::sync_with_stdio(false);
    cin.tie(0) , cout.tie(0);
    int T = read();
    while ( T -- )
    {
        init();
        for ( int i = 1 ; i <= maxn ; i ++ )
            for ( int j = 2 ; j * i <= maxn ; j ++ )
                add ( i , j * i );
        n = read();
        for ( int i = 1 ; i <= n ; i ++ ) w[i] = read();
        for ( int i = 1 ; i <= n ; i ++ ) ee[i] = read();
        for ( int i = 1 ; i <= n ; i ++ ) add ( w[i] , ee[i] );
        for ( int i = 1 ; i <= maxn ; i ++ ) if ( !dfn[i] ) tarjan(i);
        for ( int i = 1 ; i <= n ; i ++ ) if ( id[w[i]] == id[ee[i]] ) ++ ans;
        cout << ans << endl;
    }
    return 0;
}

P1653 [USACO04DEC] Cow Ski Area G

每一个合法节点和它相邻的合法四个节点连边跑一边 $t a r j a n$ 成环的相当于都能互相到达所以我们统计出度和入度为 $0$ 的点的最大值就可以知道有多少个点不能直接到达了(因为我们最终目标是消灭所有入度出度为 $0$ 的点而它们显然不能两两连边而是要从低节点连向高节点才可以消灭所以得解)

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int inf = 0x3f3f3f3f;
const int N = 500 * 500 + 5;
const int M = 500 + 5;
int read()
{
    int x = 0 , f = 1;
    char ch = getchar();
    while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
    while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
    return x * f;
}

int n , m , sumrd , sumcd , a[M][M] , rd[N] , cd[N];

int low[N] , dfn[N] , timer , in[N] , id[N] , tot , sta[N] , top;

vector<int> e[N];
inl void adde ( int u , int v ) { e[u].eb(v); }
inl int check ( int u , int v ) { return 1 <= u && u <= n && 1 <= v && v <= m; }
inl int pos ( int u , int v ) { return ( u - 1 ) * m + v; }
inl void add ( int u , int v , int p , int q )
{
    if ( !check ( p , q ) ) return;
    if ( a[u][v] >= a[p][q] ) adde ( pos ( u , v ) , pos ( p , q ) );
}

void tarjan ( int u )
{
    dfn[u] = low[u] = ++ timer;
    sta[++top] = u , in[u] = 1;
    for ( auto v : e[u] )
    {
        if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
        else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
    }
    if ( low[u] == dfn[u] )
    {
        ++ tot;
        while ( top )
        {
            int x = sta[top--];
            id[x] = tot , in[x] = 0;
            if ( x == u ) break;
        }
    }
}

signed main ()
{
    ios::sync_with_stdio(false);
    cin.tie(0) , cout.tie(0);
    m = read() , n = read();
    for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 ; j <= m ; j ++ ) a[i][j] = read();
    for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 ; j <= m ; j ++ ) add ( i , j , i , j + 1 ) , add ( i , j , i , j - 1 ) , add ( i , j , i + 1 , j ) , add ( i , j , i - 1 , j );
    for ( int i = 1 ; i <= n * m ; i ++ ) if ( !dfn[i] ) tarjan(i);
    if ( tot == 1 ) return cout << 0 << endl , 0;
    for ( int u = 1 ; u <= n * m ; u ++ ) for ( auto v : e[u] ) if ( id[u] != id[v] ) ++ cd[id[u]] , ++ rd[id[v]];
    for ( int i = 1 ; i <= tot ; i ++ ) 
        sumrd += ( rd[i] == 0 ) , sumcd += ( cd[i] == 0 );
    cout << max ( sumrd , sumcd ) << endl;
    return 0;
}

P2321 [HNOI2006] 潘多拉的宝盒

相当于是我们将合法的由 $i$ 可以升级到 $j$ 的咒语机加边然后跑 $t a r j a n$ 并求出最长链(最长升级序列)即可

对于加边我们可以同时让 $i$ 和 $j$ 从 $1$ 节点开始起跑每次让它们一起在末尾加 $0$ 或者一起在末尾加 $1$ 如果第一个点能输出而第二个点不能的话显然 $1$ 无法升级到 $2$

否则如果搜完都没有找到这样的节点那么我们可以将 $1$ 和 $2$ 连边表示升级关系

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int inf = 0x3f3f3f3f;
const int N = 500 + 5;
const int M = 500 + 5;
int read()
{
    int x = 0 , f = 1;
    char ch = getchar();
    while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
    while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
    return x * f;
}

int s , n , m , vis[N][N] , f[N] , flag , maxx;

struct node { int son[N][2] , mark[N]; } a[N];

int low[N] , dfn[N] , timer;
int scc[N] , id[N] , tot;
int in[N] , sta[N] , top;

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

void tarjan ( int u )
{
    dfn[u] = low[u] = ++ timer;
    sta[++top] = u , in[u] = 1;
    for ( auto v : e[u] )
    {
        if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
        else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
    }
    if ( low[u] == dfn[u] )
    {
        ++ tot;
        while ( top )
        {
            int x = sta[top--];
            id[x] = tot , in[x] = 0 , ++ scc[tot];
            if ( x == u ) break;
        }
    }
}

void check ( int u , int v , int x , int y )
{
    if ( vis[x][y] || !flag ) return;
    vis[x][y] = 1;
    if ( a[u].mark[x] && !a[v].mark[y] ) return flag = 0 , void();
    check ( u , v , a[u].son[x][0] , a[v].son[y][0] );
    check ( u , v , a[u].son[x][1] , a[v].son[y][1] );
}

void dp()
{
    for ( int u = tot ; u ; u -- )
        for ( auto v : ee[u] )
            f[v] = max ( f[v] , f[u] + scc[v] );
}

signed main ()
{
    ios::sync_with_stdio(false);
    cin.tie(0) , cout.tie(0);
    s = read();
    for ( int i = 1 ; i <= s ; i ++ )
    {
        n = read() , m = read();
        for ( int j = 1 ; j <= m ; j ++ ) a[i].mark[read()+1] = 1;
        for ( int j = 1 ; j <= n ; j ++ ) a[i].son[j][0] = read() + 1 , a[i].son[j][1] = read() + 1;
    }
    for ( int i = 1 ; i <= s ; i ++ )
        for ( int j = 1 ; j <= s ; j ++ )
            if ( i != j )
            {
                flag = 1 , memset ( vis , 0 , sizeof vis );
                check ( i , j , 1 , 1 );
                if ( flag ) add ( i , j ); 
            }
    for ( int i = 1 ; i <= s ; i ++ ) if ( !dfn[i] ) tarjan(i);
    for ( int u = 1 ; u <= s ; u ++ ) for ( auto v : e[u] ) if ( id[u] != id[v] ) adde ( id[u] , id[v] );
    for ( int i = 1 ; i <= tot ; i ++ ) f[i] = scc[i];
    dp();
    for ( int i = 1 ; i <= tot ; i ++ ) maxx = max ( maxx , f[i] );
    cout << maxx << endl;
    return 0;
}

P2941 [USACO09FEB] Surround the Islands

这题目是真的绕相当于是统计一个点到其他所有点的边权 $m i n$ 再乘上 $2$ (因为要走两次)

相当于是将整张图缩点然后输入时将两个岛屿之间的边权取 $m i n$ 即可

$t a r j a n$ 写错了()

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int inf = 0x3f3f3f3f;
const int N = 5e3 + 5;
int read()
{
    int x = 0 , f = 1;
    char ch = getchar();
    while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
    while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
    return x * f;
}

int n , a[N][N] , ans = inf;

int low[N] , dfn[N] , timer;
int id[N] , tot;
int sta[N] , top , in[N];

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

void tarjan ( int u )
{
    dfn[u] = low[u] = ++ timer;
    sta[++top] = u , in[u] = 1;
    for ( auto v : e[u] )
    {
        if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
        else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
    }
    if ( low[u] == dfn[u] )
    {
        ++ tot;
        while ( top )
        {
            int x = sta[top--];
            id[x] = tot , in[x] = 0;
            if ( x == u ) break;
        }
    }
}

signed main ()
{
    ios::sync_with_stdio(false);
    cin.tie(0) , cout.tie(0);
    n = read();
    memset ( a , inf , sizeof a );
    for ( int i = 1 , u , v ; i <= n ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
    for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
    for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 , val ; j <= n ; j ++ ) a[id[i]][id[j]] = min ( a[id[i]][id[j]] , read() );
    // for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 ; j <= n ; j ++ ) cout << a[i][j] << ' ';
    for ( int i = 1 ; i <= tot ; i ++ )
    {
        int sum = 0;
        for ( int j = 1 ; j <= tot ; j ++ ) sum += a[i][j];
        ans = min ( ans , sum );
    }
    cout << ( ans << 1 ) << endl; 
    return 0;
}

P3225 [HNOI2012] 矿场搭建

三倍经验：

BUSINESS - Mining your own business

Mining Your Own Business

一道好题我们可以先将所有的点双求出来我们可以为它们建一棵树（不用实际去建）

考虑统计答案如果这个连通块内没有割点那么必须建立至少两个保证一个被断之后另一个还可以发挥作用方案个数就是 $C_{n}^{2}$

如果这个连通块内有一个割点说明是叶子节点必须建立一个出口（在非割点上）因为这样如果断掉割点其他点也可以跑上面的点可以从它们父亲所能到达的其他叶子节点的出口逃跑那么我们计入答案为 $c n t + = 1$ $s u m * = (s z - 1)$

如果这个连通块内有两个以上割点说明一个点都不需要建因为如果一个割点断了另一个割点一定没断可以合并到另一个双连通分量中逃跑

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
#define int unsigned long long
const int N = 5e4 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , cases , cnt , sum , cut[N];

int low[N] , dfn[N] , timer;
int id[N] , tot;
int sta[N] , top;

vector<int> e[N] , dcc[N];
inl void add ( int u , int v ) { e[u].eb(v); }

void tarjan ( int u , int rt )
{
	int child = 0;
	dfn[u] = low[u] = ++ timer;
	sta[++top] = u;
	for ( auto v : e[u] )
		if ( !dfn[v] )
		{
			tarjan(v,rt) , low[u] = min ( low[u] , low[v] );
			if ( dfn[u] <= low[v] )
			{
				++ child , ++ tot;
				if ( u != rt || child > 1 ) cut[u] = 1;
				while ( top )
				{
					int x = sta[top--];
					dcc[tot].eb(x);
					if ( x == v ) break;	
				}
				dcc[tot].eb(u);
			}
		}
		else low[u] = min ( low[u] , dfn[v] );
}

void init()
{
	for ( int i = 1 ; i <= n ; i ++ ) e[i].clear();
	for ( int i = 1 ; i <= tot ; i ++ ) dcc[i].clear();
	memset ( cut , 0 , sizeof cut );
	timer = top = tot = cnt = 0; sum = 1;
	memset ( dfn , 0 , sizeof dfn );
}
signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	while ( 1 )
	{
		init();
		m = read(); if ( !m ) break;
		for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u ) , n = max ( n , max ( u , v ) );
		for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i,i);
		for ( int i = 1 ; i <= tot ; i ++ )
		{
			int sz = dcc[i].size() , cntcut = 0;
			for ( auto v : dcc[i] ) cntcut += cut[v];
			if ( cntcut == 1 ) ++ cnt , sum *= sz - 1;
			if ( cntcut == 0 ) cnt += 2 , sum *= sz * ( sz - 1 ) / 2;
		}
		cout << "Case " << ++ cases << ": " << cnt << ' ' << sum << endl;
	}
	return 0;
}

P3119 [USACO15JAN] Grass Cownoisseur G

显然是首先来缩点

那么对于新图上的所有边我们直接建立分层图在加边 $(u, v)$ 的时候直接加上一条 $(v, u + n)$ 的反边然后跑 $s p f a$ 即可 ( $d i j$ 貌似不能跑最长路这样的贪心策略是有问题的只能跑 $s p f a$ )

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 2e5 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , u[N] , v[N];

int low[N] , dfn[N] , timer;
int id[N] , tot , scc[N];
int in[N] , sta[N] , top;

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

void tarjan ( int u )
{
	dfn[u] = low[u] = ++timer;
	sta[++top] = u , in[u] = 1;
	for ( auto v : e[u] )
	{
		if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
		else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
	}
	if ( dfn[u] == low[u] )
	{
		++ tot;
		while ( top )
		{
			int x = sta[top--];
			in[x] = 0 , id[x] = tot , ++ scc[tot];
			if ( x == u ) break;
		}
	}
}

queue<int> q;

int dis[N];

void spfa ()
{
	memset ( in , 0 , sizeof in );
	q.push(id[1]); in[id[1]] = 1;
	while ( !q.empty() )
	{
		int u = q.front(); q.pop(); in[u] = 0;
		for ( auto v : ee[u] )
			if ( dis[v] < dis[u] + scc[v] )
			{
				dis[v] = dis[u] + scc[v]; 
				if ( !in[v] ) q.push(v) , in[v] = 1;
			}
	}
}

signed main ()
{
	// freopen ( "a.in" , "r" , stdin );
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= m ; i ++ ) u[i] = read() , v[i] = read() , add ( u[i] , v[i] );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
	for ( int i = 1 ; i <= m ; i ++ ) if ( id[u[i]] != id[v[i]] ) adde ( id[u[i]] , id[v[i]] ) , adde ( id[u[i]] + tot , id[v[i]] + tot ) , adde ( id[v[i]] , id[u[i]] + tot );
	for ( int i = 1 ; i <= tot ; i ++ ) scc[i+tot] = scc[i];
	spfa();
	cout << max ( dis[id[1]] , dis[tot+id[1]]) << endl;
	return 0;
}

P9431 [NAPC-#1] Stage3 - Jump Refreshers

一定注意在图中 $d p$ 定起点最长链的时候一定不能像普通 $t a r j a n$ 一样直接按照 $d f s$ 序来搜索而是要从该定点开始记忆化搜索(普通搜索会 $T$ )

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define pii pair<int,int>
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 3e4 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , d , s , x[N] , y[N] , f[N];

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

int dfn[N] , low[N] , timer;
int sta[N] , in[N] , top;
int id[N] , scc[N] , tot;

void tarjan ( int u )
{
	dfn[u] = low[u] = ++timer;
	in[sta[++top] = u] = 1;
	for ( auto v : e[u] ) 
	{
		if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
		else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
	}
	if ( dfn[u] == low[u] )
	{
		++ tot;
		while ( top )
		{
			int x = sta[top--];
			++ scc[tot] , in[x] = 0 , id[x] = tot;
			if ( x == u ) break;
		}
	}
}

int check ( int i , int j )
{
	return y[i] + d - y[j] >= abs ( x[i] - x[j] );
}

void init()
{
	for ( int i = 1 ; i <= n ; i ++ ) e[i].clear() , ee[i].clear();
	memset ( dfn , 0 , sizeof dfn );
	memset ( id , 0 , sizeof id );
	memset ( scc , 0 , sizeof scc );
	memset ( f , 0 , sizeof f );
	timer = tot = top = 0;
}

int dfs ( int u )
{
	if ( f[u] ) return f[u];
	f[u] = scc[u];
	for ( auto v : ee[u] )
		f[u] = max ( f[u] , scc[u] + dfs(v) );
	return f[u];
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read() , temp = read();
	while ( T -- )
	{
		init();
		n = read() , d = read() , s = read();
		for ( int i = 1 ; i <= n ; i ++ ) x[i] = read() , y[i] = read();
		for ( int i = 1 ; i <= n ; i ++ ) 
			for ( int j = 1 ; j <= n ; j ++ )
				if ( check ( i , j ) ) add ( i , j );
		for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
		for ( int u = 1 ; u <= n ; u ++ )
			for ( auto v : e[u] )
				if ( id[u] != id[v] ) adde ( id[u] , id[v] );
		cout << dfs(id[s]) << endl;
	}
	return 0;
}

[P2515 HAOI2010] 软件安装

对于一个环内的东西要么都安装要么都不安装

那么缩点之后进行树上背包即可()

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define pii pair<int,int>
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e3 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;

int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , v[N] , w[N] , rd[N] , f[N][N];

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

int dfn[N] , low[N] , timer;
int sta[N] , in[N] , top;
int id[N] , sccv[N] , sccw[N] , tot;

void tarjan ( int u )
{
    dfn[u] = low[u] = ++ timer;
    sta[++top] = u , in[u] = 1;
    for ( auto v : e[u] )
        if ( !dfn[v] ) tarjan(v) , low[u] = min ( low[u] , low[v] );
        else if ( in[v] ) low[u] = min ( low[u] , dfn[v] );
    if ( dfn[u] == low[u] )
    {
        ++ tot;
        while ( top )
        {
            int x = sta[top--];
            id[x] = tot , sccv[tot] += v[x] , sccw[tot] += w[x] , in[x] = 0;
            if ( x == u ) break;
        }
    }
}

void dfs ( int u , int ff )
{
    for ( int i = sccv[u] ; i <= m ; i ++ ) f[u][i] = sccw[u];
    for ( auto v : ee[u] )
    {
        dfs ( v , u );
        for ( int j = m ; j >= sccv[u] ; j -- )
            for ( int k = sccv[v] ; k <= j - sccv[u] ; k ++ )
                f[u][j] = max ( f[u][j] , f[u][j-k] + f[v][k] );
    }
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
    n = read() , m = read();
    for ( int i = 1 ; i <= n ; i ++ ) v[i] = read();
    for ( int i = 1 ; i <= n ; i ++ ) w[i] = read();
    for ( int i = 1 , ff ; i <= n ; i ++ ) ff = read() , add ( ff , i );
    for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
    for ( int u = 1 ; u <= n ; u ++ )
        for ( auto v : e[u] ) if ( id[u] != id[v] ) adde ( id[u] , id[v] ) , ++ rd[id[v]];
    for ( int i = 1 ; i <= tot ; i ++ ) if ( !rd[i] ) adde ( 0 , i );
    dfs ( 0 , 0 );
    cout << f[0][m] << endl;
    return 0;
}

posted @ 2023-09-18 07:26 Echo_Long 阅读(6) 评论(0) 编辑收藏举报

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