D. Vitaly and Cycle

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-2233.P5007 DDOSvoid 的疑惑2024-03-2534.P2850 [USACO06DEC] Wormholes G2024-03-2635.P1265 公路修建2024-03-2836.P1354 房间最短路问题2024-03-2937.P2168 [NOI2015] 荷马史诗2024-03-3038.P8306 【模板】字典树2024-03-3039.P1481 魔族密码2024-03-3040.P3128 [USACO15DEC] Max Flow P2024-04-0241.P5536 【XR-3】核心城市2024-04-0242.P5836 [USACO19DEC] Milk Visits S2024-04-0243.P3384 【模板】重链剖分/树链剖分2024-04-0344.P5960 【模板】差分约束2024-04-0445.P7771 【模板】欧拉路径2024-04-0546.六度分离2024-04-0847.整数区间2024-04-0848.F. Alex's whims2024-04-1249.J. 上学2024-05-1450.Game on Tree2024-05-1451.E. We Need More Bosses2024-05-1552.B. Omkar and Heavenly Tree2024-05-1653.B. Mahmoud and Ehab and the bipartiteness2024-05-1654.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1755.P6154 游走2024-05-2056.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2457.P10298 [CCC 2024 S4] Painting Roads2024-05-2458.P9650 [SNCPC2019] Escape Plan2024-05-2759.P9327 [CCC 2023 S4] Minimum Cost Roads2024-05-2960.P9026 [CCC2021 S4] Daily Commute2024-05-2961.P8724 [蓝桥杯 2020 省 AB3] 限高杆2024-06-0362.P4878 [USACO05DEC] Layout G2024-06-0463.P5663 [CSP-J2019] 加工零件2024-06-0464.P2731 [USACO3.3] 骑马修栅栏 Riding the Fences2024-06-0865.I. Disks2024-06-1766.P1351 [NOIP2014 提高组] 联合权值2024-06-2667.B. Time Travel2024-06-2968.F. Minimum Maximum Distance2024-07-1469.A. Book2024-07-1970.P1407 [国家集训队] 稳定婚姻2024-07-2071.P1991 无线通讯网2024-07-2372.P4047 [JSOI2010] 部落划分2024-07-2373.P3275 [SCOI2011] 糖果2024-07-2374.P1989 无向图三元环计数2024-07-2775.P1967 [NOIP2013 提高组] 货车运输2024-07-31

76.D. Vitaly and Cycle2024-08-01

77.P10838 『FLA - I』庭中有奇树2024-08-0778.P9751 [CSP-J 2023] 旅游巴士2024-08-1179.D. Colored Portals2024-08-21

收起

原题链接

题解

往无向图中添加至少几条边，使得图中包含奇数环？
注意是要添加少边，而不是使环小

讨论

1.0条

当且仅当原图中存在奇环，方案数为0

用bfs染色判断，即对于一个环，bfs一定会绕一圈该环，然后黑白染色判断奇偶即可

2.一条

当且仅当存在一连通图大小大于等于3

对于该连通图内，对方案数的贡献为距离为偶数的节点对数，根据黑白染色，也就是相同颜色的节点选取两个

$C_W^2+C_B^2$

3.两条

当且仅当最大的连通图大小等于2，由于有m条边，所以有m个大小为2的联通块，只需要任选一个联通块和不属于该联通块的点即可

$(n-2)\cdot m$

4.三条

联通块大小为1，也就是 $m=0$

$C_n^3$

code

#include<bits/stdc++.h>
#define ll long long
#define lb long double
#define lowbit(x) ((x)&(-x))
using namespace std;
const ll inf=1e18;
const ll mod=1e9+7;

bool flag=0;
vector<ll> G[200005];
ll color[200005]={0};
ll p1=0,p2=0;

ll dfs(ll now,ll fa,ll val)//dfs染色，子节点颜色和父节点有关
{
    color[now]=val;
    if(val==1) p1++;
    else p2++;

    ll sum=1;
    for(auto next:G[now])
    {
        if(next==fa) continue;

        if(!color[next])
        {
            sum+=dfs(next,now,-val);
        }
        else
        {
            if(color[next]==color[now])
            {
                flag=1;
                return 1;
            }
        }
    }

    return sum;
}


ll C(ll n,ll m)
{
    if(n<m) return 0;
    ll res=1;

    for(ll i=1;i<=m;i++)
    {
        res=res*(n-i+1LL);
        res=res/i;
    }
    return res;
}

void solve()
{
    ll n,m;
    cin>>n>>m;

    if(m==0)
    {
        cout<<"3 ";
        cout<<C(n,3)<<'\n';
        return;
    }

    for(ll i=1;i<=m;i++)
    {
        ll x,y;
        cin>>x>>y;

        G[x].push_back(y);
        G[y].push_back(x);
    }

    ll maxs=2;
    ll ans=0;
    for(ll i=1;i<=n;i++)
    {
        if(!color[i])
        {
            p1=0;
            p2=0;
            maxs=max(maxs,dfs(i,i,1));
            if(flag)
            {
                cout<<"0 1";
                return;
            }
            if(maxs>=3)
            {
                //printf("p1:%d  p2:%d\n",p1,p2);
                ans+=C(p1,2)+C(p2,2);
            }
        }
    }

    if(maxs==2)
    {
        cout<<"2 ";
        cout<<(n-2LL)*m<<'\n';
    }
    else
    {
        cout<<"1 ";
        cout<<ans<<'\n';
    }
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int TT=1;
    //cin>>TT;
    while(TT--) solve();
    return 0;
}