ACM 模板整理

一 .图论:

最短路：

1. Dijkstra 算法：

/*用优先队列优化，vector存储边，
输入方式：
	while (~scanf("%d%d", &m, &n) && m) {
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &a, &b, &c);
			G[a].push_back(edge(b, c));
			G[b].push_back(edge(a, c));
		}
	}
*/
#define INF 0x7fffffff
#define ll long long
#define MAX_V 1005
struct edge {
	int to, cost;
	edge() {}
	edge(int a, int b) {
		to = a;
		cost = b;
	}
};
typedef pair<int, int> P;
vector<edge> G[MAX_V];
int d[MAX_V];
int v, n, m;
void Dijkstra(int st) {
	priority_queue<P, vector<P>, greater<P>> q;
	for (int i = 0; i <= n; i++) {
		d[i] = INF;
	}
	v = -1;
	q.push(P(0, st));
	d[st] = 0;
	while (q.size()) {
		P num = q.top(); q.pop();
		v = num.second;
		if (d[v] < num.first)
			continue;
		for (int i = 0; i < G[v].size(); i++) {
			edge e = G[v][i];
			if (d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				q.push(P(d[e.to], e.to));
			}
		}
	}
}

强连通分量分解：

两个dfs 复杂度为(n + m)

//模板里的边保存一定要从0开始，不适用与从1开始
const int maxn = 1e5 + 10;
const int maxm = 5e5 + 10;
vector<int> G[maxn], rG[maxn], sc;
int n, m, topo[maxn], vis[maxn];
void addEdge(int from, int to) {
	G[from].push_back(to);
	rG[to].push_back(from);
}
void dfs(int v) {//第一遍dfs， 标记各个点的编号（后缀标记）
	vis[v] = 1;
	for (int i = 0; i < G[v].size(); i++) {
		if (!vis[G[v][i]]) dfs(G[v][i]);
	}
	sc.push_back(v);
}
void rdfs(int v, int k) {//将有向图各个边反转，进行第二遍dfs， 找出连通分量， 顺便保存缩点后的有向无环图的拓扑序
	vis[v] = 1;
	topo[v] = k;
	for (int i = 0; i < rG[v].size(); i++) {
		if (!vis[rG[v][i]]) rdfs(rG[v][i], k);
	}
}
int scc() {
	sc.clear();
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < n; i++) {
		if (!vis[i]) dfs(i);
	}
	memset(vis, 0, sizeof(vis));
	int k = 0;
	for (int i = sc.size() - 1; i >= 0; i--) { 
		if (!vis[sc[i]]) rdfs(sc[i], k++);
	}
	return k;
}