Redundant Paths POJ - 3177 kuangbin带你飞 专题九 连通图

版权声明： https://blog.csdn.net/weixin_40959045/article/details/80715914

题意

询问任意两点相互到达,要添加几条边

思路:

1. tarjan缩点
2. 注意tarjan缩点后形成的树
3. 属于同一个强连通分量的low相同
4. 竟需将树中所有入度为1的强连通分量连接即可
5. 注意应为upper_bound(deg[1] / 2) 及 (ans + 1 ) >> 1

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define MAX_V 6000
#define mem(a,b) memset(a,b,sizeof(a))
vector<int> G[MAX_V];
int ord,V;
int dfn[MAX_V],low[MAX_V];
int stk[MAX_V],top;
int instack[MAX_V];
int deg[MAX_V];
bool A[MAX_V][MAX_V];
void addedge(int from,int to)
{
  G[from].push_back(to);
  G[to].push_back(from);
}
void tarjan(int u,int fa){
  dfn[u] = low[u] = ++ord;
  instack[u] = 1;
  for (int i = 0 ; i < G[u].size() ;i++ ) {
    int v = G[u][i];
    if (v == fa) continue;
    if (dfn[v] == -1){
      tarjan(v,u);
      low[u] = min(low[u],low[v]);
    } else if (instack[v]){
      low[u] = min(dfn[v],low[u]);
    }
  }
}
void init(){
  top = ord = 0;
  for (int i = 0;i <= V ; i++) {
    G[i].clear();
  }
  mem(stk,0);
  mem(low,0);
  mem(deg,0);
  mem(instack,0);
  mem(dfn,-1);
  mem(A,0);
}
int main()
{
  int M;
  while(scanf("%d%d",&V,&M) == 2) {
    init();
    for (int i = 0;i < M ;i++){
      int u,v;
      scanf("%d%d",&u,&v);
      if (!A[u][v]){
        addedge(u,v);
        A[u][v] = A[v][u] = true;
      }
    }
    tarjan(1,-1);
    for (int u = 1;u <= V; u++ ) {
      for (int i = 0;i < G[u].size() ;i++) {
        int v = G[u][i];
        if (low[v] != low[u]){
          deg[low[v]]++;
        }
      }
    }
    int ans = 0;
    for (int i = 1;i <= V;i++){
      if (deg[i] == 1) ans++;
    }
    printf("%d\n",(ans + 1)>>1);
  }
  return 0;
}