acwing算法提高之图论--拓扑排序

1 介绍

本专题用来记录拓扑排序相关的题目。

求拓扑序列算法的关键步骤：

把入度为0的结点插入队列q。
弹出队头t（将t记录下来），遍历队头t的下一个结点，将其入度减1。操作之后，如果其值为0，则插入队列q。
重复进行步骤2，直至队列q为空。
弹出的元素组成的序列就是一个拓扑序列。

有向无环图等价于图中存在拓扑序列！

2 训练

题目1：1191家谱树

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 110;
int n;
int din[N];
vector<vector<int>> g(N);

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int t;
        while (cin >> t, t) {
            g[i].emplace_back(t);
            din[t]++;
        }
    }
    
    queue<int> q;
    for (int i = 1; i <= n; ++i) {
        if (din[i] == 0) {
            q.push(i);
        }
    }
    
    vector<int> res;
    while (!q.empty()) { //队列中存储的是，目前所有入度为0的结点
        int a = q.front();
        q.pop();
        res.emplace_back(a);
        
        for (auto b : g[a]) {
            din[b]--;
            if (din[b] == 0) {
                q.push(b);
            }
        }
    }
    
    for (auto x : res) cout << x << " ";
    cout << endl;
    
    return 0;
}

题目2：1192奖金

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 10010;
int n, m;
vector<vector<int>> g(N);
int dist[N];
int din[N];

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        int a, b;
        cin >> a >> b;
        g[b].emplace_back(a);
        din[a]++;
    }
    
    memset(dist, 0x3f, sizeof dist);
    
    queue<int> q;
    for (int i = 1; i <= n; ++i) {
        if (din[i] == 0) {
            q.push(i);
            dist[i] = 100;
        }
    }
    
    int cnt = 0;//入度为0的结点总数
    while (!q.empty()) {
        int a = q.front();
        q.pop();
        cnt++;
        
        for (int b : g[a]) {
            din[b]--;
            if (din[b] == 0) {
                q.push(b);
                dist[b] = dist[a] + 1;
            }
        }
    }
    
    if (cnt != n) puts("Poor Xed");
    else {
        int sum = 0;
        for (int i = 1; i <= n; ++i) sum += dist[i];
        cout << sum << endl;
    }
    
    return 0;
}

题目3：164可达性统计

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <bitset>
#include <vector>

using namespace std;

const int N = 30010;
int n, m;
int din[N];
bitset<N> f[N]; //f[i]表示结点i能走到的结点
vector<vector<int>> g(N);
vector<int> res;

void toposort() {
    queue<int> q;
    for (int i = 1; i <= n; ++i) {
        if (din[i] == 0) {
            q.push(i);
        }
    }
    
    while (!q.empty()) {
        int a = q.front();
        q.pop();
        res.emplace_back(a);
        
        for (auto b : g[a]) {
            din[b]--;
            if (din[b] == 0) {
                q.push(b);
            }
        }
    }
    
    return;
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        int a, b;
        cin >> a >> b;
        din[b]++;
        g[a].emplace_back(b);
    }
    
    toposort();
    
    for (int i = res.size() - 1; i >= 0; --i) {
        int a = res[i];
        f[a][a] = 1;
        for (auto b : g[a]) {
            f[a] |= f[b];
        }
    }
    
    for (int i = 1; i <= n; ++i) {
        cout << f[i].count() << endl;
    }
    
    return 0;
}

题目4：456车站分级

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 2010, M = 1000010;

int n, m;
int h[N], e[M], ne[M], w[M], idx;
int q[N], d[N];
int dist[N];
bool st[N];

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
    d[b]++;
}

void topsort() {
    int hh = 0, tt = -1;
    for (int i = 1; i <= n + m; ++i) {
        if (!d[i]) {
            q[++tt] = i;
        }
    }
    
    while (hh <= tt) {
        int t = q[hh++];
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (--d[j] == 0) {
                q[++tt] = j;
            }
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    for (int i = 1; i <= m; ++i) {
        memset(st, 0, sizeof st);
        int cnt;
        scanf("%d", &cnt);
        int start = n, end = 1;
        while (cnt--) {
            int stop;
            scanf("%d", &stop);
            start = min(start, stop);
            end = max(end, stop);
            st[stop] = true;
        }
        
        int ver = n + i;
        for (int j = start; j <= end; ++j) {
            if (!st[j]) add(j, ver, 0);
            else add(ver, j, 1);
        }
    }
    
    topsort();
    
    for (int i = 1; i <= n; ++i) dist[i] = 1;
    for (int i = 0; i < n + m; ++i) {
        int j = q[i];
        for (int k = h[j]; ~k; k = ne[k]) {
            dist[e[k]] = max(dist[e[k]], dist[j] + w[k]);
        }
    }
    
    int res = 0;
    for (int i = 1; i <= n; ++i) res = max(res, dist[i]);
    
    printf("%d\n", res);
    
    return 0;
}