LightOJ 1034 - Hit the Light Switches【强连通最小点基】

题目链接：​​http:///volume_showproblem.php?problem=1034​​

题意：有向图，选择尽量少的点，从这些点出发，使得所有点都被到达过。

思路：首先，进行强连通缩点。然后每个强连通间按照拓扑排序的思想建边。入度为0的强连通的个数即为答案。

代码：

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <sstream>
#include <malloc.h>
#include <functional>

using namespace std;

const int MAXN = 20010;
const int MAXM = 500100;

struct Edge
{
    int to, next;
}edge[MAXM];

int head[MAXM], tot;
int Low[MAXN], Dfn[MAXN], Stack[MAXN], Belong[MAXN];//Belong的值为 1 ~ scc
int Index, top;
int scc;//强连通个数
bool Instack[MAXN];
int num[MAXN];//各个强连通包含的点的个数

void addedge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void Tarjan(int u)
{
    int v;
    Low[u] = Dfn[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if (!Dfn[v])
        {
            Tarjan(v);
            if (Low[u] > Low[v])
                Low[u] = Low[v];
        }
        else if (Instack[v] && Low[u] > Dfn[v])
            Low[u] = Dfn[v];
    }
    if (Low[u] == Dfn[u])
    {
        scc++;
        do
        {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        } while (v != u);
    }
}

void solve(int N)
{
    memset(Dfn, 0, sizeof(Dfn));
    memset(Instack, false, sizeof(Instack));
    memset(num, 0, sizeof(num));
    Index = scc = top = 0;
    for (int i = 1; i <= N; i++)
    {
        if (!Dfn[i])
            Tarjan(i);
    }
}

void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}

int n, m;
int in[100010];

int main()
{
    int t, cases = 1;
    scanf("%d", &t);

    while (t--)
    {
        init();
        memset(in, 0, sizeof(in));

        scanf("%d%d", &n, &m);
        int u, v;
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d", &u, &v);
            addedge(u, v);
        }
        solve(n);
        for (int u = 1;u <= n;u++)
        {
            for (int j = head[u];j != -1;j = edge[j].next)
            {
                int v = edge[j].to;
                int t1 = Belong[u];
                int t2 = Belong[v];
                if (t1 != t2)
                    in[t2]++;
            }
        }

        int ans = 0;
        for (int i = 1;i <= scc;i++)
            if (in[i] == 0) ans++;
        printf("Case %d: %d\n", cases++, ans);

    }
    return 0;
}