POJ 1236 Network of Schools

先将图用tarjan缩点，形成无环图。然后求出其中入度为0和出度为0的点的个数ans1和ans2。

然后根据题意，入度为0的点需要添加一条边，所以第一问的结果就是ans1。这里的点都是图

的缩点。第二问取ans1和ans2大的那个。因为我们先将入度为0的点和出度为0的点连接是最

省的，剩下有点的话也得连接，所以是两者的最大值。

http://hi.baidu.com/scameeling/item/b135831094ec756771d5e815

强连通分量tarjan的介绍

/*Accepted    192K    0MS    C++    2109B    2012-07-30 10:53:03*/
#include<cstdio>
#include<cstring>
#include<cstdlib>

const int MAXN = 1 << 7;
const int MAXM = MAXN * MAXN;
int dfn[MAXN], s[MAXN], color[MAXN], low[MAXN], ins[MAXN], indgr[MAXN], outdgr[MAXN];
int first[MAXN], next[MAXM], v[MAXM], cnt, top, col, e, ans1, ans2, N;

void tarjan(int cur)
{
    int i;
    dfn[cur] = low[cur] = ++ cnt;
    s[top ++] = cur, ins[cur] = 1;
    for(i = first[cur]; i != -1; i = next[i])
    {
        if(!dfn[v[i]])
        {
            tarjan(v[i]);
            if(low[v[i]] < low[cur])
                low[cur] = low[v[i]];
        }
        else if(dfn[v[i]] < low[cur] && ins[v[i]])
            low[cur] = dfn[v[i]];
    }
    if(low[cur] == dfn[cur])
    {
        ++ col;
        for(s[top] = -1; s[top] != cur; )
            color[s[-- top]] = col, ins[s[top]] = 0;
    }
}

void cal()
{
    int i, j, zeroin, zeroout;
    cnt = top = col = 0;
    memset(ins, 0, sizeof ins);
    memset(dfn, 0, sizeof dfn);
    for(i = 1; i <= N; i ++)
        if(!dfn[i])
            tarjan(i);
    zeroin = zeroout = 0;
    for(i = 1; i <= N; i ++)
        for(j = first[i]; j != -1; j = next[j])
            if(color[i] != color[v[j]])
                ++ outdgr[color[i]], ++ indgr[color[v[j]]];
    for(i = 1; i <= col; i ++)
    {
        if(indgr[i] == 0)
            ++ zeroin;
        if(outdgr[i] == 0)
            ++ zeroout;
    }
    if( col == 1)
    {
        ans1 = 1;
        ans2 = 0;
    }
    else
    {
        ans1 = zeroin;
        ans2 = zeroin > zeroout ? zeroin : zeroout;
    }
}

void addedge(int x, int y)
{
    v[e] = y;
    next[e] = first[x], first[x] = e ++;
}

void ReadGraph()
{
    int i, j;
    e = 0;
    memset(first, -1, sizeof first);
    for(i = 1; i <= N; i ++)
    {
        for(;;)
        {
            scanf("%d", &j);
            if(0 == j) break;
            addedge(i, j);
        }
    }
}

int main()
{
    while(scanf("%d", &N) == 1)
    {
        ReadGraph();
        cal();
        printf("%d\n%d\n", ans1, ans2);
    }
    return 0;
}