文章目錄

• 題目
• 解析

題目

解析

1. 從原圖中扣出來一個樹，然後求LCA的ST表
2. 對於多餘的邊，可以對於每個邊上的點都跑一遍最短路，然後求經過這個點查詢的u,v的最短路
   OK!

// 以下是LCA模板
const int maxn = 2e5+100;
const int maxlogv = 20;
vector<int > G[maxn];
bool vis[maxn];
int id[maxn],dis[maxn];
int vs[maxn*2],depth[maxn*2];
 

int dp[maxn*2][maxlogv];
void dfs(int node,int fa,int d,int &k){
   	id[node] = k;
   	vs[k] = node;
 	depth[k++] = d;
 	// dis[node] = distance;
 	for(int i = 0;i < G[node].size(); ++i){
 		// Edge &t = G[node][i];
 		int v = G[node][i];
 		if(v == fa) continue;
 		dis[v] = dis[node]+1;
 		dfs
 
(v,node,d+1,k);
        vs[k] = node;
        depth[k++] = d;
 	}
}

void init_rmq(int n){
	
	for(int i = 0;i < n ; ++i) dp[i][0] = i;
    for(int j = 1;(1<<j) <= n; ++j){
    	for(int i = 0;i + (1<<j)-1 < n; ++i){
    		if(depth[dp[i][j-1]]< depth[dp[i+(1<<(j-1))][j-1]])
    			dp[
 
i][j] = dp[i][j-1];
    		else
    			dp[i][j] = dp[i+(1<<(j-1))][j-1];

    	}
    }
}
int query(int l,int r){
	int k = 0;
	while((1<<(k+1)) <= r-l+1) k++;
 	if(depth[dp[l][k]] < depth[dp[r-(1<<k)+1][k]])
 		return dp[l][k];
 	else 
 		return dp[r-(1<<k)+1][k];
}
int lca(int u,int v){
	return vs[query(min(id[u],id[v]),max(id[u],id[v]))];
}
void init(int n){
	int k = 0;
	dfs(0,-1,0,k);
	init_rmq(2*n-1);
}


// 本題相關
vector<int > G1[maxn];
int d[maxn];
int ans[maxn];
int qu[maxn],qv[maxn];
int n,m,q;
set<int> s;
void dfs1(int u,int fa){
	vis[u]  = 1;
	for(int i = 0;i < G1[u].size(); ++i){
		int v = G1[u][i];
		if(v == fa) continue;
		if(vis[v]){
          s.insert(v);
          s.insert(u);
          continue;
		}
		dfs1(v,u);
		G[u].Pb(v);
	}
}

void Bfs(int s){
   
   
   memset(vis,0,sizeof(vis));
   rep(i,0,n) d[i] = INF;
   queue<int> Q;
   Q.push(s);
   d[s] = 0;
   vis[s] = 1;
   while(!Q.empty()){
  	 int u  = Q.front(); Q.pop();
     for(int i = 0,v;i <G1[u].size(); ++i){
     	v = G1[u][i];
     	if(vis[v]) continue;
     	d[v] = d[u]+1;
     	vis[v] = 1;
     	Q.push(v);
     }
   }
   rep(i,0,q){
   	ans[i] = min(ans[i],d[qu[i]]+d[qv[i]]);
   }
}

int main(void){
   
    scanf("%d%d",&n,&m);
    	
    	for(int i = 0;i < m; ++i){
    		int u,v;
    		scanf("%d%d",&u,&v);
    		u--,v--;
    		G1[u].push_back(v);
    		G1[v].push_back(u);
    	}
        dfs1(0,-1);
    	init(n);
    	scanf("%d",&q);
    	rep(i,0,q){
    		scanf("%d%d",&qu[i],&qv[i]);
    		qu[i]--,qv[i]--;
    		ans[i] = dis[qu[i]]+dis[qv[i]]-2*dis[lca(qu[i],qv[i])];
    	}
    	for(auto c: s){
        	Bfs(c);
        }
        rep(i,0,q){
        	printf("%d\n",ans[i]);
        }
	  return 0;
}



``