裸題：1191. 家譜樹

1191. 家譜樹 - AcWing題庫

#include <iostream>
#include <cstring>
using namespace std;

const int N = 110, M = 10010;
int h[N], e[M], ne[M], idx;
int d[N], q[N], hh, tt = -1;
int n, m;

void add(int x, int y)
{
    e[idx] = y, ne[idx] = h[x], h[x] = idx ++ ;
}

void topsort()
{
    for (int i = 1; i <= n; ++ i )
        if (!d[i]) q[ ++ tt ] = i;
    while (tt >= hh )
    {
        int x = q[hh ++ ];
        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            if (-- d[y] == 0) q[++ tt] = y;
        }
    }
}

int main()
{
    memset(h, -1, sizeof(h));
    scanf("%d", &n);
    for (int x = 1; x <= n; ++ x )
    {
        int y;
        while (scanf("%d", &y), y)
        {
            add(x, y);
            d[y] ++ ;
        }
    }
    
    topsort();
    for (int i = 0; i <= tt; ++ i )
        printf("%d ", q[i]);
    return 0;
}

差分約束+拓?fù)渑判颍?192. 獎(jiǎng)金

1192. 獎(jiǎng)金 - AcWing題庫

由于圖中所有邊權(quán)都是正數(shù)，可以直接使用topsort求解差分約束問題
根據(jù)題意，要求一個(gè)最小值，使用最長路求解，轉(zhuǎn)化題目的條件：$A>=B+1$與$x_i>=x_0+100$
$x_0$為一個(gè)虛擬源點(diǎn)，向每個(gè)點(diǎn)連了一條權(quán)值為100的邊

若圖中存在環(huán)，topsort的隊(duì)列長度將小于n，因?yàn)榄h(huán)的起點(diǎn)無法進(jìn)入隊(duì)列
先用topsort判斷圖中是否存在環(huán)，若不存在，根據(jù)拓?fù)湫虮闅v圖，求解最長路

#include <iostream>
#include <cstring>
using namespace std;

const int N = 10010, M = 30010;
int h[N], e[M], ne[M], w[M], idx;
int q[N], d[N], hh, tt = -1;
int dis[N];
int n, m;

void add(int x, int y, int d)
{
    e[idx] = y, ne[idx] = h[x], w[idx] = d, h[x] = idx ++ ;
}

bool topsort()
{
    q[ ++ tt ] = 0;
    while (tt >= hh)
    {
        int x = q[hh ++ ];
        for (int i = h[x]; i != -1; i = ne[i] )
        {
            int y = e[i];
            if ( -- d[y] == 0) q[ ++ tt ] = y;
        }
    }
    return tt == n;
}

int main()
{
    memset(h, -1, sizeof(h));
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; ++ i )
    {
        int x, y;
        scanf("%d%d", &x, &y);
        add(y, x, 1);
        d[x] ++ ;
    }
    
    for (int i = 1; i <= n; ++ i ) add(0, i, 100), d[i] ++ ;
    if (!topsort()) puts("Poor Xed");
    else 
    {
        for (int k = 0; k <= tt; ++ k )
        {
            int x = q[k];
            for (int i = h[x]; i != -1; i = ne[i])
            {
                int y = e[i];
                dis[y] = max(dis[y], dis[x] + w[i]);
            }
        }
        int sum = 0;
        for (int i = 1; i <= n; ++ i ) sum += dis[i];
        printf("%d\n", sum);
    }
    
    return 0;
}

debug：最后的遍歷沒有按照拓?fù)湫?/p>

for (int x = 0; x <= tt; ++ x )
{
	for (int i = h[x]; i != -1; i = ne[i])
	{
		int y = e[i];
		dis[y] = max(dis[y], dis[x] + w[i]);
	}
}

集合+拓?fù)湫颍?64. 可達(dá)性統(tǒng)計(jì)

[164. 可達(dá)性統(tǒng)計(jì) - AcWing題庫](https://www.acwing.com/problem/content/description/166/

從集合的角度思考，$f(i)$表示i這個(gè)點(diǎn)能到達(dá)的所有點(diǎn)，i首先能到達(dá)自己，其次能達(dá)到$f(j_1),f(j_2),...,f(j_n)$，假設(shè)i與n個(gè)點(diǎn)直接相連
那么要求$f(i)$，就必須求出$f(j_1),f(j_2),...,f(j_n)$，即拓?fù)渑判蛑形挥趇之后的所有點(diǎn)的$f(j)$
所以這題先拓?fù)渑判?，再根?jù)拓?fù)渑判虻哪嫘颍?span id="n5n3t3z" class="katex--inline">$f(i)$
如何表示集合$f(i)$？用STL的容器bitset，假設(shè)圖中有N個(gè)點(diǎn)，那么bitset的長度為N，每個(gè)點(diǎn)都用一個(gè)bitset記錄其集合，1表示i能遞達(dá)這個(gè)點(diǎn)，0表示不能遞達(dá)
那么$f(i)=f(j_1)\cap f(j_2)\cap ...\cap f(j_n)$

關(guān)于bitset的使用，bitset之間支持|=運(yùn)算，count()輸出bitset中1的個(gè)數(shù)

#include <iostream>
#include <cstring>
#include <bitset>
using namespace std;

const int N = 30010, M = N;
int h[N], e[M], ne[M], idx;
int q[N], d[N], hh, tt = -1;
bitset<N> f[N];
int n, m;

void add(int x, int y)
{
    e[idx] = y, ne[idx] = h[x], h[x] = idx ++ ;
}

void topsort()
{
    for (int i = 1; i <= n; ++ i )
        if (!d[i]) q[ ++ tt ] = i;
    while (tt >= hh)
    {
        int x = q[hh ++ ];
        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            if ( -- d[y] == 0) q[ ++ tt ] = y;
        }
    }
}

int main()
{
    memset(h, -1, sizeof(h));
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; ++ i )
    {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y);
        d[y] ++ ;
    }
    topsort();
    
    for (int i = tt; i >= 0; -- i )
    {
        int x = q[i];
        f[x][x] = 1;
        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            f[x] |= f[y];
        }
    }
    for (int i = 1; i <= n; ++ i ) printf("%d\n", f[i].count());

    return 0;
}

差分約束+拓?fù)湫颍?56. 車站分級(jí)

456. 車站分級(jí) - AcWing題庫

分析題意：對(duì)于每一條路線，未經(jīng)過的站點(diǎn)的等級(jí)一定小于經(jīng)過的站點(diǎn)等級(jí)，并且最低的站點(diǎn)等級(jí)為1級(jí)
題目要求所有等級(jí)劃分中的最少等級(jí)數(shù)，用最長路求最小值。將以上條件轉(zhuǎn)換成差分約束中的兩個(gè)條件：$B>=A+1$, $x_i>=x_0+1$
$x_0$為虛擬源點(diǎn)，通過$x_0$能到達(dá)圖中的所有點(diǎn)，那么就一定能遞達(dá)所有邊

由于每條路線路都會(huì)建立n * m條邊，極端情況下可能會(huì)爆空間，所以考慮如何優(yōu)化
一條路徑中未經(jīng)過的站點(diǎn)將向經(jīng)過的站點(diǎn)連接一條權(quán)值為1的邊，一共n * m條，由于這些邊的權(quán)值相同，可以在這些邊中創(chuàng)建一個(gè)虛擬點(diǎn)v，未經(jīng)過的點(diǎn)分別向v連一條權(quán)值為0的邊，v向經(jīng)過的點(diǎn)分別連接一條權(quán)值為1的邊。這樣，從未經(jīng)過的點(diǎn)到經(jīng)過的點(diǎn)的權(quán)值和依然為1，但是需要建立的邊數(shù)為n + m，此時(shí)的邊數(shù)在極端情況下也不會(huì)爆空間

#include <iostream>
#include <cstring>
using namespace std;

const int N = 2010, M = 1e6 + 10;
int h[N], e[M], ne[M], w[M], idx;
int d[N], q[N], hh, tt = -1;
bool st[N]; int dis[N];
int n, m;

void add(int x, int y, int d)
{
    e[idx] = y, ne[idx] = h[x], w[idx] = d, h[x] = idx ++ ;
}

void topsort()
{
    for (int i = 1; i <= n + m; ++ i ) 
        if (!d[i]) q[ ++ tt ] = i;
    while (tt >= hh)
    {
        int x = q[hh ++ ];
        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            if (-- d[y] == 0) q[ ++ tt ] = y;
        }
    }
}

int main()
{
    memset(h, -1, sizeof(h));
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++ i )
    {
        memset(st, false, sizeof(st));
        int t, start = n, end = 0;
        scanf("%d", &t);
        while (t -- )
        {
            int x;
            scanf("%d", &x);
            st[x] = true;
            start = min(start, x), end = max(end, x);
        }
        int v = n + i;
        for (int i = start; i <= end; ++ i )
        {
            if (st[i]) add(v, i, 1), d[i] ++ ;
            else add(i, v, 0), d[v] ++ ;
        }
    }
    
    topsort();
    for (int i = 1; i <= n; ++ i ) dis[i] = 1;
    for (int i = 1; i <= tt; ++ i )
    {
        int x = q[i];
        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            dis[y] = max(dis[y], dis[x] + w[i]);
        }
    }
    int res = 0;
    for (int i = 1; i <= n; ++ i ) res = max(res, dis[i]);
    printf("%d\n", res);
    
    return 0;
}

debug：w[M]寫成了w[N]，又是這樣，然后debug了半天，了文章來源地址http://www.zghlxwxcb.cn/news/detail-650546.html

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