题意
给定一个无向图，问删掉点i，图中相连的有序对数。(pair<x, y> , x != y);
求每个点对应的答案
思路
首先我们可以发现，如果这个点不是割点，那么答案就是n-1，如果是割点，就要考虑子树中的联通块。
可以用tarjan，O（n）的复杂度
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 来了老弟 ＼('ㅅ') > ⌒ヽ / へ＼ / / ＼＼ ﾚ ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / 'ノ ) Lﾉ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; struct E{ int v,nxt; }edge[1000009]; int head[maxn],gtot; ll ans[maxn]; void addedge(int u,int v){ edge[gtot].v = v; edge[gtot].nxt = head[u]; head[u] = gtot++; } int dp[maxn],dfn[maxn],low[maxn]; int tot = 0; int n,m; void dfs(int u,int fa){ dfn[u] = low[u] = ++tot; dp[u] = 1; int sum = 0; for(int i=head[u]; ~i; i = edge[i].nxt){ int v = edge[i].v; if(dfn[v] == 0){ dfs(v, u); low[u] = min(low[u], low[v]); dp[u] += dp[v]; if(low[v] >= dfn[u]) { ans[u] += 1ll*sum * dp[v]; sum = sum + dp[v]; } } else if(v != fa){ low[u] = min(low[u], dfn[v]); } } if(fa != -1)ans[u] = ans[u] + 1ll*(n-sum-1)*sum; } int main(){ memset(head, -1, sizeof(head)); scanf("%d%d", &n, &m); for(int i=1; i<=m; i++){ int u,v; scanf("%d%d", &u, &v); addedge(u,v); addedge(v,u); } dfs(1, -1); for(int i=1; i<=n; i++) { printf("%lld ", (ans[i] + 1ll*(n-1)) * 2ll); } return 0; }