题解 P1656 炸铁路

题目

枚举割掉一条边，然后用并查集看是否为会有两个（或者多个）不联通点。

#include <bits/stdc++.h>
using namespace std;
#define N 5010
#define ll long long

template <class T>
inline void read(T& a){
	T x = 0, s = 1;
	char c = getchar();
	while(!isdigit(c)){ if(c == '-') s = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + (c ^ '0'); c = getchar(); }
	a = x * s;
	return ;
}

int n, m;

struct edge{
  int u, v, next;
} t[N << 1];
int head[N / 10]; 

int bian = 0;
inline void addedge(int u, int v){
  t[++bian] = (edge){u, v, head[u]}, head[u] = bian;
  t[++bian] = (edge){v, u, head[v]}, head[v] = bian;
  return ; 
}

vector < pair<int, int> > ans; 
vector < pair<int, int> > G; 
int su, sv; 

bool vis[N / 10]; 

int fa[N]; 

int find_fa(int x){
  return x == fa[x] ? x : fa[x] = find_fa(fa[x]); 
}

inline void merge(int u, int v){
  int fau = find_fa(u), fav = find_fa(v);
  if(fau != fav) fa[fau] = fav; 
  return ; 
}

inline void clean(){
  memset(fa, 0, sizeof(fa));
  memset(vis, 0, sizeof(vis));
  for(int i = 1; i <= n; i++) fa[i] = i; 
  return ; 
}

void dfs(int now){
  if(vis[now]) return ;
  vis[now] = 1;
  for(int i = head[now]; i; i = t[i].next){
    int v = t[i].v;
    if((su == now && sv == v) || (su == v && sv == now)) continue ; 
    merge(now, v);
    dfs(v); 
  }
  return ; 
}

int main(){
  read(n), read(m);
  for(int i = 1; i <= m; i++){
    int x, y;
    read(x), read(y);
    addedge(x, y); 
    G.push_back(make_pair(x, y)); 
  }
  for(int i = 0; i < G.size(); i++){
    clean(); 
    su = G[i].first;
    sv = G[i].second; 
    bool flag = 0;  
    for(int j = 1; j <= n; j++)
      if(!vis[j]) dfs(j); 
    for(int j = 1; j <= n; j++)
      for(int h = j + 1; h <= n; h++){
        int fau = find_fa(j), fav = find_fa(h);
        if(fau != fav){
          flag = 1;
          break; 
        }
      }
    if(flag) ans.push_back(make_pair(min(su, sv), max(su, sv))); 
  }
  sort(ans.begin(), ans.end());
  for(int i = 0; i < ans.size(); i++){
    printf("%d %d\n", ans[i].first, ans[i].second); 
  }

	return 0;
}