P2746 [USACO5.3] 校园网Network of Schools

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2024-03-07

17.P2746 [USACO5.3] 校园网Network of Schools2024-03-08

18.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2024-07-3176.D. Vitaly and Cycle2024-08-0177.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

把奶牛看成点，赠送列表关系看成有向边，这样这道题就成了对强连通分量缩点，然后找出这个新图中入度为零的点有几个，出度为零的点有几个

code

#include<bits/stdc++.h>
using namespace std;
vector<int> G[105];
int len=0,cnt=0;
int belong[105]={0};
int in[105]={0},out[105]={0};
int in_st[105]={0},low[105]={0},order[105]={0};
stack<int> q;
void sd(int now)//有向图不需要判断父节点
{
    order[now]=low[now]=++len;
    q.push(now);
    in_st[now]=1;
    for(auto next:G[now])
    {
        if(!order[next])
        {
            sd(next);
            low[now]=min(low[now],low[next]);
        }
        else if(in_st[next]) low[now]=min(low[now],order[next]);//判断是否在栈内实际上是判断这个点是否是这一轮遍历的，不然会错误的更新
    }

    if(order[now]==low[now])
    {
        int next;
        cnt++;
        do
        {
            next=q.top();
            q.pop();
            in_st[next]=0;
            belong[next]=cnt;
        }while(next!=now);
    }
}
int main()
{
    int n,x;
    cin>>n;
    for(int i=1;i<=n;i++) while(cin>>x&&x) G[i].push_back(x);//有向图建边
    for(int i=1;i<=n;i++)if(!order[i])sd(i);//如果这个点没有缩过，那就缩点
    for(int i=1;i<=n;i++)//遍历所有边，更新边两端归属不同的缩点的入度和出度
    for(auto j:G[i])
    {
        int x=belong[i],y=belong[j];
        if(x!=y)
        {
            out[x]++;
            in[y]++;
        }
    }

    int ins=0,outs=0;
    for(int i=1;i<=cnt;i++)
    {
        ins+=(in[i]==0);
        outs+=(out[i]==0);
    }
    if(cnt==1)puts("1\n0");//如果只有一个点
    else cout<<ins<<endl<<max(ins,outs)<<endl;
    return 0;
}