cf1335F. Robots on a Grid

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题解

这题还蛮不好写的

首先对这个基环树森林的每个环求出大小，每个环上找一个代表点

一棵基环树上的所有点按照其到环上代表点的距离模环大小的余数分组

每种余数只能取一个

然后我贪心先取黑的

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct Graph
{
    int etot, head[maxn], to[maxe], next[maxe], w[maxe];
    void clear(int N)
    {
        for(int i=1;i<=N;i++)head[i]=0;
        etot=0;
    }
    void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}
    #define forp(_,__) for(auto p=__.head[_];p;p=__.next[p])
}G;
struct Trajan_SCC
{
    int dfn[maxn], low[maxn], vis[maxn], scc[maxn], tim, scc_cnt;
    stack<int> s;
    void clear(int n)
    {
        for(int i=1;i<=n;i++)vis[i]=0;
        tim=scc_cnt=0;
    }
    void dfs(Graph &G, int pos)
    {
        int p;
        dfn[pos]=low[pos]=++tim;
        s.emplace(pos);
        vis[pos]=1;
        for(p=G.head[pos];p;p=G.next[p])
        {
            if(vis[G.to[p]]==0)dfs(G,G.to[p]);
            if(vis[G.to[p]]==1)low[pos]=min(low[pos],low[G.to[p]]);
        }
        if(dfn[pos]==low[pos])
        {
            int x;
            scc_cnt++;
            do
            {
                x=s.top(), s.pop();
                scc[x]=scc_cnt;
                vis[x]=2;
            }
            while(x!=pos);
        }
    }
    void run(Graph &G, int n)
    {
        for(int i=1;i<=n;i++)if(!vis[i])dfs(G,i);
    }
}tarjan;
vector<string> col, dir;
ll root[maxn], fa[maxn], n, m, vis[maxn], dis[maxn], len[maxn];
set<ll> white[maxn], black[maxn];
ll id(ll x, ll y)
{
    return x*m+y;
}
ll get_size(ll x)
{
    ll cnt=1, t=fa[x];
    while(t!=x)cnt++,t=fa[t];
    return cnt;
}
void dfs(ll x)
{
    vis[x]=1;
    if(vis[fa[x]]==0)dfs(fa[x]),root[x]=root[fa[x]];
    else if(vis[fa[x]]==1)root[x]=x;
    else root[x]=root[fa[x]];
    dis[x]=dis[fa[x]]+1;
    vis[x]=2;
}
int main()
{
    ios::sync_with_stdio(false);
    ll T, i, j;
    cin>>T;
    while(T--)
    {
        col.clear(), dir.clear();
        cin>>n>>m;
        rep(i,0,n-1)
        {
            string s;
            cin>>s;
            col.emb(s);
        }
        rep(i,0,n-1)
        {
            string s;
            cin>>s;
            dir.emb(s);
        }
        rep(i,0,n*m-1)vis[i]=0;
        rep(i,0,n-1)rep(j,0,m-1)
        {
            ll x=id(i,j);
            switch( dir[i][j] )
            {
                case 'U':
                fa[x]=id(i-1,j);break;
                case 'D':
                fa[x]=id(i+1,j);break;
                case 'L':
                fa[x]=id(i,j-1);break;
                case 'R':
                fa[x]=id(i,j+1);break;
            }
        }
        rep(i,0,n*m-1)if(vis[i]==0)dfs(i);
        rep(i,0,n*m-1)white[i].clear(), black[i].clear();
        rep(i,0,n*m-1)if(root[i]==i)len[i]=get_size(i);
        rep(i,0,n-1)rep(j,0,m-1)
        {
            ll x=id(i,j);
            if(col[i][j]=='1')white[root[x]].em( dis[x]%len[root[x]] );
            else black[root[x]].em( dis[x]%len[root[x]] );
        }
        ll ans1=0, ans2=0;
        rep(i,0,n*m-1)
        {
            ans2+=black[i].size();
            for(auto x:black[i])white[i].erase(x);
            ans1+=white[i].size();
        }
        cout<<ans1+ans2<<' '<<ans2<<'\n';
    }
    return 0;
}