POJ 1236 Network of Schools

 

大意：

第一问：最小的数量需要收到复制软件从而使得所有学校都有软件。（You are to write a program that computes the minimal number of schools that must receive a copy of the new software）

第二问：建立最小数量的“路”，使得任意的学校两两可达。

思路：第一问，“缩点”之后判断入度为0的强连通分量。第二问，与Hdu 2767很相似，求连接最小的路使得图变为强连通图。判断入度0与出度为0的最大值。

CODE：

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

#define MAXN 50010
#define MAXM 100010

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

int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int ind[MAXN], outd[MAXN];

int n, m;
int cnt;
int scnt, top, tot;

void init()
{
    cnt = 0;
    scnt = top = tot = 0;
    memset(dfn, 0, sizeof(dfn));
    memset(first, -1, sizeof(first));
    memset(ins, 0, sizeof(ins));
    memset(ind, 0, sizeof(ind));
    memset(outd, 0, sizeof(outd));
}

void read_graph(int u, int v)
{
    edge[cnt].v = v;
    edge[cnt].next = first[u];
    first[u] = cnt++;
}

void dfs(int u)
{
    int v;
    dfn[u] = low[u] = ++tot;
    ins[u] = 1;
    stack[top++] = u;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        v = edge[e].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u])
    {
        scnt++;
        do
        {
            v = stack[--top];
            belong[v] = scnt;
            ins[v] = 0;
        }while(v != u);
    }
}

void Tarjan()
{
    for(int v = 1; v <= n; v++) if(!dfn[v])
        dfs(v);
}

void solve()
{
    Tarjan(); //Tarjan
    if(scnt == 1) { printf("1\n0\n"); return ;} //本身是强连通图 
    for(int u = 1; u <= n; u++)
    {
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(belong[u] != belong[v]) ind[belong[v]]++, outd[belong[u]]++;
        }
    }
    int max1 = 0, max2 = 0, ans = 0;
    for(int i = 1; i <= scnt; i++)
    { 
        if(!outd[i]) max1++;
        if(!ind[i]) max2++;
    }
    printf("%d\n", max2);
    printf("%d\n", max1 > max2 ? max1:max2);
}

int main()
{
    while(~scanf("%d", &n))
    {
        init();
        for(int u = 1; u <= n; u++)
        {
            int v;
            while(scanf("%d", &v) && v)
            {
                read_graph(u, v);
            }
        }
        solve();
    }
    return 0;
}