hihor 學習日記:hiho一下 第五十三週(邊的雙連通分量)
阿新 • • 發佈:2018-12-30
http://hihocoder.com/contest/hiho53/problem/1
求出邊的雙聯通分量
與求割點與割邊類似,求邊的雙聯通分量找的是橋,有m橋,那麼就有m+1的分組。
而分組有點類似割邊,如果一個分組中存在割邊那麼這個分組就不是分組,所以這張圖的割邊就是分組的邊界,不包括在分組中,而low[u] == cur[u]正是割邊所在的點,這是利用棧和DFS,把u後面入棧的點全部劃分成同一分組,這樣就寫完了
AC程式碼:
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int Mod = 1e9 + 7;
const int maxn = 1e5 + 5;
const double eps = 0.00000001;
const int INF = 0x3f3f3f3f;
struct Edge{
int v, next;
}edge[maxn << 1];
int tot, head[maxn];
int parent[maxn], vis[maxn], low[maxn], cur[maxn], trace[maxn];
int top, sta[maxn], cnt;
void init() {
cnt = top = tot = 0;
memset(head, -1, sizeof(head));
memset(parent, -1, sizeof(parent));
memset(vis, 0, sizeof(vis));
}
void addEdge(int u, int v) {
edge[tot].v = v;
edge[tot].next = head[u];
head[u] = tot ++;
edge[tot].v = u;
edge[tot].next = head[v];
head[v] = tot ++;
}
void DFS (int u) {
static int counter = 0;
int child = 0;
vis[u] = 1;
cur[u] = low[u] = ++counter;
sta[++ top] = u;
for (int i = head[u]; i + 1; i = edge[i].next) {
int v = edge[i].v;
if(!vis[v]) {
child ++;
parent[v] = u;
DFS(v);
low[u] = min(low[u], low[v]);
if(cur[u] < low[v]) {
cnt ++;
//printf("bridge: %d %d\n", u, v);
}
}else if(v != parent[u])
low[u] = min(low[u], cur[v]);
}
if(low[u] >= cur[u]) {
int tt = top;
int minn = u;
while(sta[tt] != u) {
minn = min(minn, sta[tt]);
tt --;
}
while(sta[top] != u) {
trace[sta[top]] = minn;
top --;
}
if(sta[top] == u) {
trace[u] = minn;
top --;
}
}
}
int main()
{
init();
int N, M;
cin >> N >> M;
while(M --) {
int u, v;
cin >> u >> v;
addEdge(u, v);
}
DFS(1);
cout << cnt + 1<< endl;
for (int i = 1; i <= N; i ++)
cout << trace[i] << " ";
cout << endl;
return 0;
}