目录
- [1 介绍](#1 介绍)
- [2 训练](#2 训练)
- [3 参考](#3 参考)
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，
cpp
#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，
cpp
#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，
cpp
#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，
cpp
#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;
}