AcWing 396. 矿场搭建 题解(tarjan割点)

AcWing 396. 矿场搭建
懒得写了，直接照搬大佬的题解

大佬题解原地址

#include<bits/stdc++.h>

using namespace std;

typedef unsigned long long ULL;

const int N = 1010, M = 510;

int h[N], e[M], ne[M], idx;
int n ,m;
int low[N], dfn[N], timep;
vector<int>dcc[N];  //储存分割块内都有那些点
bool cut[N];  //记录是不是割点 
int root;
int dcc_cnt;  //记录分割块数量 
int stk[N];
int top;
 
void add(int a, int b){
    
	e[idx] = b;
	ne[idx] = h[a];
	h[a] = idx ++ ;
} 

void tarjan(int u){
    
	low[u] = dfn[u] = ++ timep;
	stk[ ++ top] = u;
	
	//如果是孤点 
	if(u == root && h[u] == -1){
    
		dcc_cnt ++ ;  //分割块数量++
		dcc[dcc_cnt].push_back(u);
		return ; 
	}
	
	//如果不是孤点
	int cnt = 0;
	for(int i = h[u]; ~i; i = ne[i]){
    
		int j = e[i];
		if(!dfn[j]){
    
			tarjan(j);
			low[u] = min(low[u], low[j]);
			if(dfn[u] <= low[j]){
    
				cnt ++ ;
				if(u != root || cnt > 1) cut[u] = true;  //是个割点 
				++ dcc_cnt;
				int y;
				do{
    
					y = stk[top -- ];
					dcc[dcc_cnt].push_back(y); 
				}while(y != j);  //因为要保留u给其他包含u的分割块，所以不将u弹出 
				dcc[dcc_cnt].push_back(u);  //这个分割块内还是要加入u节点 
			}			
		}
		else low[u] = min(low[u], dfn[j]);
	} 
}

int main()
{
    
	int T = 1;
	while(scanf("%d", &m), m){
    
		for(int i = 1; i <= dcc_cnt; i ++ ) dcc[i].clear();  //分割块的数量是从1开始用的 
		
		n = idx = timep = dcc_cnt = top = 0;
		memset(h, -1, sizeof h);
		memset(dfn, 0, sizeof dfn);
		memset(cut, 0, sizeof cut);
		
		while(m -- ){
    
			int a, b;
			cin>>a>>b;
			n = max(n, max(a, b));
			add(a, b);
			add(b, a);
		}
		
		for(root = 1; root <= n; root ++ ){
    
			if(!dfn[root])
				tarjan(root);
		}
		
		//遍历每一个分割块
		int res = 0;
		ULL num = 1;  //记录方案数 
		for(int i = 1 ;i <= dcc_cnt; i ++ ){
    
			int cnt = 0;  //记录分割点数 
			for(int j = 0; j < dcc[i].size(); j ++ ){
    
				if(cut[dcc[i][j]]) cnt ++ ;
			} 
			
			if(cnt == 0){
      //如果没有割点 
				if(dcc[i].size() > 1) res += 2, num *= dcc[i].size() * (dcc[i].size() - 1) / 2;  //这不是个孤点 
				else res ++ ; 
			}
			else if(cnt == 1) res ++ , num *= dcc[i].size() - 1;
		} 
		printf("Case %d: %d %llu\n", T ++, res, num);
	}
	return 0;
}