题解【LOJ10094】「一本通 3.5 练习 2」消息的传递

题面

很裸的强连通分量。

Tarjan 缩点后，统计入度为 (0) 的点的个数，直接输出即可。

#include <bits/stdc++.h>

using namespace std;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 1003, M = N * N;

int n, m;
int head[N], headc[N], ver[M], nxt[M], tot;
int dfn[N], low[N], tim, stk[N], topp, scc_cnt, id[N];
bool in_stk[N];
int in[N];

inline void add(int h[], int u, int v)
{
    ver[++tot] = v, nxt[tot] = h[u], h[u] = tot;
}

void Tarjan(int u)
{
    dfn[u] = low[u] = ++tim, stk[++topp] = u, in_stk[u] = true;
    for (int i = head[u]; i; i = nxt[i])
    {
        int v = ver[i];
        if (!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (in_stk[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u])
    {
        int y = -1;
        ++scc_cnt;
        do
        {
            y = stk[topp--];
            in_stk[y] = false;
            id[y] = scc_cnt;
        } while (y != u);
    }
}

int main()
{
    n = gi();
    for (int i = 1; i <= n; i+=1)
        for (int j = 1; j <= n; j+=1)
        {
            int x = gi();
            if (x) add(head, i, j);
        }
    for (int i = 1; i <= n; i+=1) if (!dfn[i]) Tarjan(i);
    for (int u = 1; u <= n; u+=1)
        for (int i = head[u]; i; i = nxt[i])
        {
            int v = ver[i];
            if (id[u] != id[v])
                add(headc, id[u], id[v]), ++in[id[v]];
        }
    int ans = 0;
    for (int i = 1; i <= scc_cnt; i+=1)
        if (!in[i])
            ++ans;
    cout << ans << endl;
	return 0;
}