NC51269 Network of Schools

题目链接

https://ac.nowcoder.com/acm/problem/51269

题意

给你一张有向图，问最少加几条边使得图上的点属于同一个强连通分量。

思路

缩点后入度为0的点和出度为0的点取最大值， 特判缩点完只有一个点的情况。

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const int INF = 0x3f3f3f3f;
struct node{
    int to, next;
} edge[maxn];
stack<int> s;
int tot, sc, cnt;//sc为缩点后总点数
int dfn[maxn], low[maxn];
int scc[maxn], in[maxn], out[maxn];
int head[maxn], vis[maxn];
int sz[maxn];
void add(int from, int to){
    edge[++cnt].to = to;edge[cnt].next = head[from];head[from] = cnt;
}
void ins(int u, int v){add(u, v); add(v, u);}
void Tarjan(int u){
    dfn[u] = low[u] = ++tot;
    vis[u] = 1;
    s.push(u);
    for(int i = head[u];i != -1;i = edge[i].next){
        int v = edge[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]){
        sc++;
        int v;
        do{
            v = s.top();s.pop();
            scc[v] = sc;
            vis[v] = 0;
            sz[sc]++;
        }while(v != u);
    }
}
int main()
{
    std::ios::sync_with_stdio(false);
    int n;
    cin >> n;
    memset(head, -1, sizeof(head));
    for(int i = 1;i <= n;i++){
        int x;
        while(cin >> x){
            if(!x) break;
            add(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 != -1;j = edge[j].next){
            int u = scc[i];
            int v = scc[edge[j].to];
            if(u != v) in[v]++, out[u]++;
        }
    }
    int cnt1 = 0, cnt2 = 0;
    for(int i = 1;i <= sc;i++){
        if(in[i] == 0) cnt1++;
        if(out[i] == 0)cnt2++;
    }
    if(sc == 1) cout << 1 << "\n" << 0 << endl;
    else cout << cnt1 << "\n" << max(cnt1, cnt2) << endl;
    return 0;
}