poj 1236 Network of Schools ： 求需要添加多少条边成为强连通图 tarjan O(E)

/**
problem: http://poj.org/problem?id=1236

缩点后入度为0的点的总数为需要发放软件的学校个数
缩点后出度为0的点的总数和入度为0的点的总数的最大值为需要增加的传输线路的条数（头尾相接）
特别的，当图为强连通图时，发放软件的学校个数为1， 增加线路为0
**/
#include<stdio.h>
#include<stack>
#include<algorithm>
using namespace std;

class Graphics{
    const static int MAXN = 105;
    const static int MAXM = 100005;
private:
    struct Edge{
        int to, next;
        bool bridge;
    }edge[MAXM];
    struct Point{
        int dfn, low, color;
    }point[MAXN];
    int sign, dfnNum, colorNum, sumOfPoint, first[MAXN];
    bool vis[MAXN];
    stack<int> stk;
    void tarjan(int u){
        point[u].dfn = ++dfnNum;
        point[u].low = dfnNum;
        vis[u] = true;
        stk.push(u);
        for(int i = first[u]; i != -1; i = edge[i].next){
            int to = edge[i].to;
            if(!point[to].dfn){
                tarjan(to);
                point[u].low = min(point[to].low, point[u].low);
                if(point[to].low > point[u].dfn){
                    edge[i].bridge = true;
                }
            }else if(vis[to]){
                point[u].low = min(point[to].dfn, point[u].low);
            }
        }
        if(point[u].dfn == point[u].low){
            vis[u] = false;
            point[u].color = ++colorNum;
            while(stk.top() != u){
                vis[stk.top()] = false;
                point[stk.top()].color = colorNum;
                stk.pop();
            }
            stk.pop();
        }
    }
public:
    void clear(int n){
        sign = dfnNum = colorNum = 0;
        for(int i = 1; i <= n; i ++){
            first[i] = -1;
            vis[i] = 0;
        }
        sumOfPoint = n;
        while(!stk.empty()) stk.pop();
    }
    void addEdgeOneWay(int u, int v){
        edge[sign].to = v;
        edge[sign].bridge = false;
        edge[sign].next = first[u];
        first[u] = sign ++;
    }
    void addEdgeTwoWay(int u, int v){
        addEdgeOneWay(u, v);
        addEdgeOneWay(v, u);
    }
    void tarjanAllPoint(){
        for(int i = 1; i <= sumOfPoint; i ++){
            if(!point[i].dfn)
                tarjan(i);
        }
    }
    pair<int, int> getAns(){
        int ans = 0, ans2 = 0;
        int *indegree = new int[sumOfPoint+1];
        int *outdegree = new int[sumOfPoint+1];
        for(int i = 1; i <= sumOfPoint; i ++){
            indegree[i] = 0;
            outdegree[i] = 0;
        }
        tarjanAllPoint();
        for(int i = 1; i <= sumOfPoint; i ++){
            for(int j = first[i]; j != -1; j = edge[j].next){
                int to = edge[j].to;
                if(point[to].color != point[i].color){
                    outdegree[point[i].color] ++;
                    indegree[point[to].color] ++;
                }
            }
        }
        for(int i = 1; i <= colorNum; i ++){
            if(!indegree[i]){
                ans ++;
            }
            if(!outdegree[i]){
                ans2 ++;
            }
        }
        delete []indegree; delete []outdegree;
        if(colorNum == 1){
            return make_pair(1, 0);
        }else{
            return make_pair(ans, max(ans, ans2));
        }
    }
}graph;

int main(){
    int n;
    scanf("%d", &n);
    graph.clear(n);
    for(int i = 1; i <= n; i ++){
        int a;
        while(scanf("%d", &a) && a){
            graph.addEdgeOneWay(i, a);
        }
    }
    pair<int, int> ans = graph.getAns();
    printf("%d
%d
", ans.first, ans.second);
    return 0;
}

 ps:

这两句话是不一样的，在有向图中存在非环边还是非桥的情况