Question: Undirected graph, how many strong connected components after deleting two points.
Problem solution: strong connected component + cut point
First consider deleting a point, and then consider finding the cut point of the remaining graph. The added connected block is represented by add_block, plus the original number of connected blocks to find the maximum.
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const int MAXN = 10010;
const int MAXM = 100010;
struct Edge {
int to, next;
bool cut;//是否为桥的标记
}edge[MAXM];
int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN];
int Index, top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int bridge;
void addedge(int u, int v) {
edge[tot].to = v; edge[tot].next = head[u]; edge[tot].cut = false;
head[u] = tot++;
}
void Tarjan(int u, int pre, int x) {
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int son = 0;
int pre_cnt = 0; //处理重边，如果不需要可以去掉
for (int i = head[u]; ~i; i = edge[i].next) {
v = edge[i].to;
if (v == x) continue;
if (v == pre && pre_cnt == 0) {
pre_cnt++; continue; }
if (!DFN[v]) {
son++;
Tarjan(v, u, x);
if (Low[u] > Low[v]) Low[u] = Low[v];
//割点
//一个顶点 u 是割点，当且仅当满足 (1) 或 (2)
// (1) u 为树根，且u 有多于一个子树。
// (2) u 不为树根，且满足存在 (u,v) 为树枝边 (或称父子边，
// 即 u 为 v 在搜索树中的父亲)，使得 DFS(u)<=Low(v)
if (u != pre && Low[v] >= DFN[u]) {
//不是树根
cut[u] = true;
add_block[u]++;
}
}
else if (Low[u] > DFN[v])
Low[u] = DFN[v];
}
//树根，分支数大于 1
if (u == pre && son > 1) cut[u] = true;
if (u == pre) add_block[u] = son - 1;
Instack[u] = false;
top--;
}
int cnt;
void solve(int N, int x) {
memset(DFN, 0, sizeof(DFN));
memset(Instack, false, sizeof(Instack));
memset(add_block, 0, sizeof(add_block));
memset(cut, false, sizeof(cut));
cnt = 0;
Index = top = 0;
bridge = 0;
for (int i = 1; i <= N; i++)
if (!DFN[i] && i != x) Tarjan(i, i, x), cnt++;
}
void init() {
tot = 0;
memset(head, -1, sizeof(head));
}
int n, m, u, v;
int main() {
while (~scanf("%d%d", &n, &m)) {
init();
for (int i = 1; i <= m; i++) {
scanf("%d%d", &u, &v);
u++, v++;
addedge(u, v);
addedge(v, u);
}
int ans = 0;
for (int i = 1; i <= n; i++) {
solve(n, i);
for (int j = 1; j <= n; j++) {
if (i == j) continue;
ans = max(ans, cnt + add_block[j]);
}
}
printf("%d\n", ans);
}
return 0;
}