AcWing 1175. 最大半连通子图

思路

• 1.先利用tarjan把所有强连通分量算出来，将图变为一个有向无环图
• 2.对这个有向无环图建图，并且除去重边
• 3.对于一个有向无环图可以用dp的方式f[i]表示的是以i点为终点的最大连通子图的点的个数，g[i]表示这种情况的方案数
• 4.接下来就是枚举一下所有有向无环图的点，对于每一个点枚举边，进行dp，最后求f最大值即可

代码

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long LL;

const int N = 100010, M = 2000010;

int n, m, mod;
int h[N], hs[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
stack<int> stk;
bool in_stk[N];
int id[N], scc_cnt, scc_size[N];
int f[N], g[N];

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

void tarjan(int u){
    dfn[u]=low[u]=++timestamp;
    stk.push(u);
    in_stk[u]=true;
    for(int i=h[u];~i;i=ne[i]){
        int j=e[i];
        if(!dfn[j]){
            tarjan(j);
            low[u]=min(low[u],low[j]);
        }
        else{
            if(in_stk[j]){
                low[u]=min(low[u],low[j]);
            }
        }
    }
    if(dfn[u]==low[u]){
        ++ scc_cnt;
        int y;
        do{
            y=stk.top();
            stk.pop();
            in_stk[y] = false;
            id[y] = scc_cnt;
            scc_size[scc_cnt] ++ ;
        }while(y!=u);
    }
    
}
int main()
{
    memset(h, -1, sizeof h);
    memset(hs, -1, sizeof hs);

    scanf("%d%d%d", &n, &m, &mod);
    while (m -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(h, a, b);
    }

    for (int i = 1; i <= n; i ++ )
        if (!dfn[i])
            tarjan(i);

    unordered_set<LL> S;    // (u, v) -> u * 1000000 + v
    for (int i = 1; i <= n; i ++ )
        for (int j = h[i]; ~j; j = ne[j])
        {
            int k = e[j];
            int a = id[i], b = id[k];
            LL hash = a * 1000000ll + b;//转化为hash值来存下每条边
            if (a != b && !S.count(hash))//判重
            {
                add(hs, a, b);
                S.insert(hash);
            }
        }

    for (int i = scc_cnt; i; i -- )
    {
        if (!f[i])//初始化
        {
            f[i] = scc_size[i];
            g[i] = 1;
        }
        for (int j = hs[i]; ~j; j = ne[j])
        {
            int k = e[j];
            if (f[k] < f[i] + scc_size[k])
            {
                f[k] = f[i] + scc_size[k];
                g[k] = g[i];
            }
            else if (f[k] == f[i] + scc_size[k])
                g[k] = (g[k] + g[i]) % mod;
        }
    }

    int maxf = 0, sum = 0;
    for (int i = 1; i <= scc_cnt; i ++ )
        if (f[i] > maxf)
        {
            maxf = f[i];
            sum = g[i];
        }
        else if (f[i] == maxf) sum = (sum + g[i]) % mod;

    printf("%d\n", maxf);
    printf("%d\n", sum);

    return 0;
}