【洛谷 SP2878】Knights of the Round Table（双联通分量）

先放这吧，没时间写，明天再补

“明天到了”

题目链接

题意：求不在任何奇环内的点的数量。

Tarjan求点双联通分量，然后再染色判断是不是二分图就好了。

只是不懂为什么Tarjan求双联通分量时要用栈保存点对，希望大佬留言帮助。

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
inline int min(int a, int b){
    return a > b ? b : a;
}
inline int max(int a, int b){
    return a > b ? a : b;
}
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
const int MAXN = 1010;
const int MAXM = 1000010;
int n, m;
int a, b;
int s[MAXN][MAXN];
struct Edge{
    int next, to;
}e[MAXM << 1];
int head[MAXN], num;
inline void Add(int from, int to){
    e[++num].to = to;
    e[num].next = head[from];
    head[from] = num;
}
int dfn[MAXN], low[MAXN], ID, color[MAXN], can[MAXN], belong[MAXN], top, ans, cnt;
struct point{
    int u, v;
}stack[MAXN << 2];
vector <int> dcc[MAXN];
bool Judge(int u, int Color, int now){
    color[u] = Color;
    for(int i = head[u]; i; i = e[i].next){
       if(belong[e[i].to] != now) continue;
       if(!color[e[i].to])
         if(Judge(e[i].to, 3 - Color, now)) return true;
         else;
       else
         if(color[e[i].to] == color[u])
           return true;
    }
    return false;
}
void Tarjan(int u, int fa){
    dfn[u] = low[u] = ++ID;
    for(int i = head[u]; i; i = e[i].next){
       if(!dfn[e[i].to]){
         stack[++top] = (point){ u, e[i].to };
         Tarjan(e[i].to, u);
         low[u] = min(low[u], low[e[i].to]);
         if(low[e[i].to] >= dfn[u]){
           dcc[++cnt].clear();
           point now;
           do{
             now = stack[top--];
             if(belong[now.u] != cnt) belong[now.u] = cnt, dcc[cnt].push_back(now.u);
             if(belong[now.v] != cnt) belong[now.v] = cnt, dcc[cnt].push_back(now.v);
           }while(now.u != u || now.v != e[i].to);
         }
       }
       else if(dfn[e[i].to] < dfn[u] && e[i].to != fa)
         stack[++top] = (point){ u, e[i].to }, low[u] = min(low[u], dfn[e[i].to]);
    }
}
int main(){
    while(233){
      n = read(); m = read();
      if(!n && !m) break;
      memset(s, 0, sizeof s);
      memset(dfn, 0, sizeof dfn);
      memset(low, 0, sizeof low);
      memset(can, 0, sizeof can);
      memset(head, 0, sizeof head);
      memset(color, 0, sizeof color);
      memset(belong, 0, sizeof belong);
      ID = ans = top = num = cnt = 0;
      for(int i = 1; i <= m; ++i){
         a = read(); b = read();
         s[a][b] = s[b][a] = 1;
      }
      for(int i = 1; i < n; ++i)
         for(int j = i + 1; j <= n; ++j)
            if(!s[i][j])
              Add(i, j), Add(j, i);
      for(int i = 1; i <= n; ++i)
         if(!dfn[i])
           Tarjan(i, 0);
      for(int i = 1; i <= cnt; ++i){
         memset(color, 0, sizeof color);
         for(int j = 0; j < dcc[i].size(); ++j)
            belong[dcc[i][j]] = i;
         if(Judge(dcc[i][0], 1, i)){
           for(int j = 0; j < dcc[i].size(); ++j)
              can[dcc[i][j]] = 1;
         }
      }
      for(int i = 1; i <= n; ++i)
         if(!can[i])
           ++ans;
      printf("%d\n", ans);
    }
    return 0;
}