洛谷 2002_消息扩散_强连通分量

题目描述

有n个城市，中间有单向道路连接，消息会沿着道路扩散，现在给出n个城市及其之间的道路，问至少需要在几个城市发布消息才能让这所有n个城市都得到消息。

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思路

用tarjian求强连通分量，输出入度为0的个数即可

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#include <stdio.h>#include <stack>#include <string>#include <cstring>using namespace std;#define maxn 600000#define fill(x, y) memset(x, y, sizeof(x))struct edge{    int from, to, next;}e[maxn];int ls[maxn], f[maxn], n, m, dfn[maxn], low[maxn], bcnt = 0, instack[maxn], dindex = 0, belong[maxn], in[maxn];stack<int> s;void tarjan(int now){    dindex++;    dfn[now] = dindex;    low[now] = dindex;    s.push(now);    instack[now] = 1;    for (int i = ls[now]; i; i = e[i].next)    {        if (dfn[e[i].to] == 0)        {            tarjan(e[i].to);            if (low[now] > low[e[i].to]) low[now] = low[e[i].to];        }        else        {            if (instack[e[i].to] && dfn[e[i].to] < low[now])                low[now] = dfn[e[i].to];        }    }    if (dfn[now] == low[now])    {        bcnt++;        int j;        do        {            j = s.top();            s.pop();            instack[j] = 0;            belong[j] = bcnt;        }        while(now != j);    }}void work(){    for (int i = 1; i <= n; i++)        if (dfn[i] == 0) tarjan(i);}int main(){    scanf("%d%d", &n, &m);    for (int i = 1; i <= m; i++)    {        int x, y;        scanf("%d%d", &x, &y);        e[i] = (edge) {x, y, ls[x]};        ls[x] = i;    }    work();    for (int i = 1; i <= m; i++)        if (belong[e[i].from] != belong[e[i].to])            in[belong[e[i].to]]++;    int ans = 0;    for (int i = 1; i <= bcnt; i++)        if (in[i] == 0) ans++;    printf("%d\n", ans);}