55pts超时求调

//O(nm)，无向连通图 
#include <bits/stdc++.h>
#define int long long
#define pp pair<int, int>
#define ppp pair<pair<int, int>, int>
#define pppp pair<pair<int, int>, pair<int, int> >
#define p_q priority_queue
#define gvi int, vector<int>, greater<int>
#define gvp pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> >
using namespace std;
void fast(){ ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); }
int min(int x, int y){ return x < y ? x : y; }
int max(int x, int y){ return x > y ? x : y; }
const int INF=0x3f3f3f3f, M=505;
int n, m, cnt;
struct node{
	int num, u, v, w;	//存边 
}g[M*M];
int dis[M][M];	//第m天任意两点之间的距离 
stack<int> s;
void floyd(int x, int y){
	for (int i=1; i<=n; i++)
	for (int j=1; j<=n; j++)
	dis[i][j]=min(dis[i][j], dis[i][x] + dis[x][j]);
	
	for (int i=1; i<=n; i++)
	for (int j=1; j<=n; j++)
	dis[i][j]=min(dis[i][j], dis[i][y] + dis[y][j]);
	return ;
}
int calc(){	//计算总成本 
	cnt=0;
	for (int i=1; i<=n; i++)
	for (int j=i+1; j<=n; j++)
	cnt+=dis[i][j];
	return cnt*2;
}
void init(){
	for (int mid=1; mid<=n; mid++)
	for (int i=1; i<=n; i++)
	for (int j=1; j<=n; j++)
	dis[i][j]=min(dis[i][j], dis[i][mid] + dis[mid][j]);
	s.push(calc());
	return ;
} 
signed main(){
	fast();
	memset(dis, INF, sizeof dis);	//初始状态赋最大值 
	cin >> n >> m;
	for (int i=1; i<=m; i++){
		g[i].num=i;
		cin >> g[i].u >> g[i].v; 
		dis[g[i].u][g[i].v] = dis[g[i].v][g[i].u] = 1; 
	} 
	init();	//先floyd跑一遍第m天的情况 
	for (int i=m; i>1; i--){	//倒序 
		dis[g[i].u][g[i].v] = dis[g[i].v][g[i].u] = 0; 
		//往前每一天都有一条道路变成无需通行费 
		floyd(g[i].u, g[i].v); s.push(calc());
	}
	while (!s.empty()){	//输出答案 
		cout << s.top() << '\n';
		s.pop();
	}
	return 0;
}