F. Microcycle

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-0429.P3957 [NOIP2017 普及组] 跳房子2024-03-0630.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0732.P1379 八数码难题2024-03-0733.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-0940.P4147 玉蟾宫2024-03-1041.P2866 [USACO06NOV] Bad Hair Day S2024-03-1042.P2032 扫描2024-03-1043.P1884 [USACO12FEB] Overplanting S2024-03-1044.D. Lonely Mountain Dungeons2024-03-12

45.F. Microcycle2024-03-12

46.E. Increasing Subsequences__22024-03-1347.D. Valiant's New Map2024-03-1448.Array Repetition2024-03-1449.F. Rudolf and Imbalance2024-03-1450.D. Exam in MAC2024-03-1551.G. Rudolf and Subway2024-03-1552.C. Ehab and Path-etic MEXs2024-03-1753.A. String Transformation 12024-03-1754.D. Solve The Maze2024-03-1855.B. Array Fix2024-03-1856.B. Equal XOR2024-03-1857.F. Maximum White Subtree2024-03-1958.P1652 圆2024-03-2059.P2181 对角线2024-03-2060.P1325 雷达安装2024-03-2061.P3194 [HNOI2008] 水平可见直线2024-03-2062.B - Make Many Triangles2024-03-2363.[ARC172A] Chocolate2024-03-2564.C. Theofanis' Nightmare2024-03-2765.D. Birthday Gift2024-03-2766.C. Arrow Path2024-03-2767.P1265 公路修建2024-03-2868.P1284 三角形牧场2024-03-2869.Kirill and Mushrooms2024-03-2870.C. Sasha and the Casino2024-03-2871.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0172.P5536 【XR-3】核心城市2024-04-0273.P5836 [USACO19DEC] Milk Visits S2024-04-0274.P4391 [BOI2009] Radio Transmission 无线传输2024-04-0475.P3435 [POI2006] OKR-Periods of Words2024-04-0476.P5960 【模板】差分约束2024-04-0477.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0578.E. Long Inversions2024-04-0979.C. Deep Down Below2024-04-0980.C. Rings2024-04-0981.F. Unfair Game2024-04-1082.G. GCD on a grid2024-04-1083.C. Inversion Graph2024-04-1084.Increase Subarray Sums2024-04-1085.D. Cross Coloring2024-04-1286.M. Triangle Construction2024-04-1287.D2. Set To Max (Hard Version)2024-04-1288.C. Array Game2024-04-1289.F. Alex's whims2024-04-1290.Queue Sort2024-04-1291.P1908 逆序对2024-04-1392.P5094 [USACO04OPEN] MooFest G 加强版2024-04-1593.P5149 会议座位2024-04-1594.P10337 [UESTCPC 2024] 操作序列2024-04-1595.P10330 [UESTCPC 2024] 黑白珠串2024-04-1596.P10252 线性变换2024-04-1697.P3901 数列找不同2024-04-1698.P10288 [GESP样题 八级] 区间2024-04-1699.P5677 [GZOI2017] 配对统计2024-04-17

收起

原题链接

题解

原题可以理解为 $\to$ 找出含有最小边权的边双连通分量，然后找出那个边对应的两个节点的第二条路

本题我学会了：
1.代码命名风格要含义清晰，不然很容易搞混，包括变量，自定义函数
2.建边的时候尽量用链表式，因为dalao都在用，看题解方便一点

code

#define ll long long
#include<bits/stdc++.h>
using namespace std;
ll depth=0;
stack<ll> q;
ll dfn[200010]={0},low[200010]={0};
ll cnt=0;
ll belong[200010]={0};
ll last[400010]={0};
struct node
{
    int from,to,val,head;
}edge[400010];
void add(int from,int to,int val,int num)
{
    edge[num].to=to;
    edge[num].from=from;
    edge[num].val=val;
    edge[num].head=last[from];
    last[from]=num;
}

inline void read(ll &x) {
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}


void flss(ll now,ll fa)
{
    dfn[now]=low[now]=++depth;

    q.push(now);
    for(int i=last[now];i;i=edge[i].head)
    {
        ll to=edge[i].to;
        if(to==fa) continue;
        if(!dfn[to])
        {
            flss(to,now);
            low[now]=min(low[now],low[to]);
        }
        else low[now]=min(low[now],dfn[to]);
    }

    if(low[now]==dfn[now])
    {
        cnt++;
        ll x;
        do
        {
            x=q.top();
            q.pop();
            belong[x]=cnt;
        }while(x!=now);
    }
}


stack<int> st;
int vis[200005]={0};
int dfs(ll now,ll fa,ll ends)
{
    st.push(now);
    vis[now]=1;

    if(now==ends)
    {
        cout<<st.size()<<endl;
        while(st.size())
        {
            cout<<st.top()<<" ";
            vis[st.top()]=0;
            st.pop();
        }

        puts("");
        return 1;
    }

    for(int i=last[now];i;i=edge[i].head)
    {
        ll to=edge[i].to,val=edge[i].val;
        if(to==fa||vis[to]||belong[to]!=belong[now]) continue;
        if (dfs(to,now,ends)) return 1;
    }
    st.pop();
    //vis[now]=0;导致我#19TLE的罪魁祸首
    return 0;
}


int main()
{
    ll t;
    read(t);
    while(t--)
    {
        ll n,m;
        read(n); read(m);
        for(ll i=1;i<=n;i++)
        {
            dfn[i]=0;
            low[i]=0;
            last[i]=0;
            vis[i]=0;
            belong[i]=0;
        }
        cnt=0;
        depth=0;
        for(ll i=1;i<=m;i++)
        {
            ll x,y,w;
            read(x); read(y); read(w);
            add(x,y,w,i);
            add(y,x,w,i+m);
        }

        for(ll i=1;i<=n;i++) if(!dfn[i]) flss(i,i);
        ll ans=2e17;
        for(ll i=1;i<=m;i++)
        {
            ll x=edge[i].from,y=edge[i].to,val=edge[i].val;
            if(belong[x]==belong[y]) ans=min(ans,val);
        }

        write(ans); putchar(' ');
        for(ll i=1;i<=m;i++)
        {
            ll x=edge[i].from,y=edge[i].to,val=edge[i].val;
            if(belong[x]==belong[y]&&val==ans)
            {
                dfs(x,y,y);
                break;
            }
        }
    }
    return 0;
}