Network of Schools(POJ1326+有向图进行缩点）

题目链接：http://poj.org/problem?id=1236

题目：

题意：对于n个学校，对于一个系统传给某个学校，那么他会传给他得支援学校。从第二开始，每行给你多个数字，表示第i个学校可以支援这些学校，以0结尾。问你一个新软件至少要给多少个学校，如果任意传给某个学校都能传给其他学校需要建多少条支援关系。

思路：tarjan进行缩点，重新建图，对新建得有向无环图统计一下出度和入度，第一问答案就是入度为0的数量，第二问则是max（入度为0的个数，出度为0的个数）。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 10007;
const int maxn = 100 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int n, tot, x, cnt, num, top;
int head[maxn], c[maxn];
int in[maxn], out[maxn], vis[maxn];
int dfn[maxn], low[maxn], stc[maxn];

void init() {
    tot = cnt= num = top = 0;
    memset(c, 0, sizeof(c));
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(vis, 0, sizeof(vis));
    memset(stc, 0, sizeof(stc));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(head, -1, sizeof(head));
}

struct edge {
    int v, next;
}ed[maxn*maxn];

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

void tarjan(int x) {
    dfn[x] = low[x] = ++num;
    stc[++top] = x, vis[x] = 1;
    for(int i = head[x]; ~i; i = ed[i].next) {
        int y = ed[i].v;
        if(!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        } else if(vis[y]) {
            low[x] = min(low[x], low[y]);
        }
    }
    if(dfn[x] == low[x]) {
        cnt++; int y;
        do {
            y = stc[top--]; vis[y] = 0;
            c[y] = cnt;
        }while(x != y);
    }
}

int main() {
    //FIN;
    scanf("%d", &n);
    init();
    for(int i = 1; i <= n; i++) {
        while(1) {
            scanf("%d", &x);
            if(x == 0) break;
            addedge(i, x);
        }
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i]) {
            tarjan(i);
        }
    }
    for(int i = 1; i <= n; i++) {
        for(int j = head[i]; ~j; j = ed[j].next) {
            int y = ed[j].v;
            if(c[i] == c[y]) continue;
            out[c[i]]++;
            in[c[y]]++;
        }
    }
    if(cnt == 1) {
        printf("1
0
");
        return 0;
    }
    int ans1 = 0, ans2 = 0;
    for(int i = 1; i <= cnt; i++) {
        if(in[i] == 0) ans1++;
        if(out[i] == 0) ans2++;
    }
    printf("%d
%d
", ans1, max(ans1, ans2));
    return 0;
}