POJ 1236 Network of Schools強連通分量縮點

/*相當經典的題目了. 題的描述有點兒長, 我就不想再描述了...題中的任務A, 其本質就是求這個有向圖中所有強連通分量中, 入度為0的個數. 原理還是挺好理解的:入度不為0的強連通分量, 顯然可以由其他強連通分量達到. 也就是說, 要想讓所有學校都拿到軟體,需要且只需要從入度為0的那些強連通分量出發即可.題中的任務B稍微複雜些., 意思是增加多少條邊, 可以使整個圖變成強連通的. 這時需要分別統計入度為0和出度為0的點, 設它們的個數分別是a, b. 首先, 我們可以在入度為0和出度為0的點之間連邊. 之後,如果入度 / 出度為0的點有剩餘, 就把它們和其他點連邊. 因此, 需要連的邊條數為 max(a, b).只有一個聯通塊的時候，輸出零。因為不需要添加邊*/#include <cstdio>#include <iostream>using namespace std;#define maxn 105#define maxm 3000int top;//棧頂位置int Bcnt;//強連通分量編號int Index;//時間順序int DFN[maxn];//時間戳記int LOW[maxn];int belong[maxn];//頂點i屬於哪個強連通分量int Stack[maxn];//棧int instack[maxn];//是否在棧內int n, m;struct node{    int to;    int next;} edge[maxm];int head[maxn];bool Judge[maxn];bool Judge2[maxn];int ansi;void init(){    fill(head, head + n + 1, -1);    fill(DFN, DFN + n + 1, 0);    fill(Judge, Judge + n + 1, true);    fill(Judge2, Judge2 + n + 1, true);    ansi = 0;    top = 0;    Bcnt = 0;    Index = 0;}void add(int a, int b){    edge[ansi].to = b;    edge[ansi].next = head[a];    head[a] = ansi++;}void read(){    int a;    for (int i = 1; i <= n; i++)    {        while (scanf("%d", &a) && a)            add(i, a);    }}void tarjan(int i){    int j, k;    DFN[i] = LOW[i] = ++Index;    instack[i] = true;    top++;    Stack[top] = i;    for (k = head[i]; k != -1; k = edge[k].next)    {        j = edge[k].to;        if (!DFN[j])//j未訪問，用dfn值標記是否已訪問過        {            tarjan(j);            if (LOW[j] < LOW[i])                LOW[i] = LOW[j];        }        else if (instack[j] && DFN[j] < LOW[i])            LOW[i] = DFN[j];    }    if (DFN[i] == LOW[i]) //dfn和low相等，遞迴列印強連通分量    {        Bcnt++;//強連通分量編號        do        {            j = Stack[top--];            instack[j] = false;            belong[j] = Bcnt;        }        while (j != i);    }}void judge1(){    for (int i = 1; i <= n; i++)        for (int k = head[i]; k != -1; k = edge[k].next)        {            if (belong[i] != belong[edge[k].to])                Judge[belong[i]] = false;        }}void judge2(){    for (int i = 1; i <= n; i++)        for (int k = head[i]; k != -1; k = edge[k].next)        {            if (belong[i] != belong[edge[k].to])                Judge2[belong[edge[k].to]] =false;        }}void solve(){    init();    read();    for (int i = 1; i <= n; i++)        if (!DFN[i])            tarjan(i);    judge1();    judge2();    int s1=0;//所有出度為零的    int s2=0;//所有入度為零的    for (int i = 1; i<=Bcnt; i++)    {        if (Judge[i])s1++;        if(Judge2[i])s2++;    }    printf("%d\n",s2);    int ss;    if(Bcnt==1)ss=0;    else ss=max(s1,s2);    printf("%d\n",ss);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif    while (scanf("%d", &n) != EOF)    {        solve();    }}