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;

}