Knights of the Round Table POJ

题目：https://vjudge.net/problem/POJ-2942

一个月前花了大精力做的，现在又忘完啦！

题目大意

有奇数个骑士围着桌子坐，相互憎恨的骑士不能坐在一起，问你有几个骑士没处坐。

分析

1. 当两个骑士可以相邻坐，把他俩连边，那么就可以得到一个以上的无向图。
2. 由性质得，一个简单奇圈上的所有节点必然属于同一个双连通分量，那么题目就转化为求不在任一奇圈上的节点个数。
3. 首先明确，二分图是没有奇圈的，所以只用关注那些不是二分图的BCC，给不是二分图的BCC标记即可。

代码：

/******************************************************************** * File Name: Knights_of_the_round_table.cpp * Author: Sequin * mail: [e-mail] * Created Time: 五  9/22 18:01:36 2017 *************************************************************************/#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <ctype.h>#include <set>#include <vector>#include <cmath>#include <bitset>#include <algorithm>#include <climits>#include <string>#include <list>#include <cctype>#include <cstdlib>#include <fstream>#include <sstream>using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define pi acos(-1.0)#define pii pair<int,int>#define ll long long#define MAX 10000005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)#define maxn 1002ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}struct Edge{    int u, v;};int odd[maxn];int color[maxn];int pre[maxn], iscut[maxn], bccno[maxn], dfs_clock, bcc_cnt;vector<int> G[maxn], bcc[maxn];stack<Edge> S;bool bipartite(int u, int b) {    for(int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if(bccno[v] != b) {            continue;        }        if(color[v] == color[u]) {            return false;        }        if(!color[v]) {            color[v] = 3 - color[u];            if(!bipartite(v, b)) {                return false;            }        }    }    return true;}int dfs(int u, int fa) {    int lowu = pre[u] = ++dfs_clock;    int child = 0;    for(int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        Edge e = (Edge) {u, v};        if(!pre[v]) {            S.push(e);            child++;            int lowv = dfs(v, u);            lowu = min(lowu, lowv);            if(lowv >= pre[u]) {                iscut[u] = true;                bcc_cnt++;                bcc[bcc_cnt].clear();                W(true) {                    Edge x = S.top();                    S.pop();                    if(bccno[x.u] != bcc_cnt) {                        bcc[bcc_cnt].push_back(x.u);                        bccno[x.u] = bcc_cnt;                    }                    if(bccno[x.v] != bcc_cnt) {                        bcc[bcc_cnt].push_back(x.v);                        bccno[x.v] = bcc_cnt;                    }                    if(x.u == u && x.v == v) {                        break;                    }                }            }        }        else if(pre[v] < pre[u] && v != fa) {            S.push(e);            lowu = min(lowu, pre[v]);        }    }    if(fa < 0 && child == 1) {        iscut[u] = 0;    }    return lowu;}void find_bcc(int n) {    MEM(pre, 0);    MEM(iscut, 0);    MEM(bccno, 0);    for(int i = 1; i <= n; i++) {        if(!pre[i]) {            dfs(i, -1);        }    }}int mm[maxn][maxn];int main() {    std::ios::sync_with_stdio(false);    int n ,m;    while(cin >> n >> m && n + m) {        for(int i = 1; i <= n; i++) {            G[i].clear();        }        MEM(mm, 0);        for(int i = 0; i < m; i++) {            int a, b;            cin >> a >> b;            mm[a][b] = 1;            mm[b][a] = 1;        }        for(int i = 1; i <= n; i++) {            for(int j = i + 1; j <= n; j++) {                if(!mm[i][j]) {                    G[i].push_back(j);                    G[j].push_back(i);                }            }        }        dfs_clock = 0;        bcc_cnt = 0;        find_bcc(n);        MEM(odd, 0);        for(int i = 1; i <= bcc_cnt; i++) {            MEM(color, 0);            for(int j = 0; j < bcc[i].size(); j++) {                bccno[bcc[i][j]] = i;            }            int u = bcc[i][0];            color[u] = 1;            if(!bipartite(u, i)) {                for(int j = 0; j < bcc[i].size(); j++) {                    odd[bcc[i][j]] = 1;                }            }        }        int ret = n;        for(int i = 1; i <= n; i++) {            if(odd[i]) {                ret--;            }        }        cout << ret << endl;    }}