CF626F. Bear and Fair Set

Solution

单走一个网络流。
先对余数$0..4$分别建一个点，从$S->0..4$分别连容量$n/5$的边。

对于每一个限制，相当于把$[0,b]$分成了若干个小区间，并且可以求得每个小区间中数的个数，然后由$0..4$的点$t$分别向每个小区间$[l,r]$，连一条容量为${\sum }_{l\le i\le r}[imod5=t]$，即区间内模$5$余数为$t$的数的个数。

最后从每个区间向$T$连一条容量为区间长度的边。

跑网络流判断是否满流即可。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=300005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
PR a[MAXN];
queue<int> que;
struct enode{int nxt,v,c; } e[MAXN];
int dist[MAXN],start[MAXN],head[MAXN],nodenum=1,n,R,m,S,T,N;
int get(int l,int r,int k)
{int ans=0;for (int i=l;i<=r;i++) ans+=(i%5==k);return ans;
}
void add(int u,int v,int c)
{
//	cout<<u<<" "<<v<<" "<<c<<endl;if (!c) return;e[++nodenum]=(enode){head[u],v,c},head[u]=nodenum;e[++nodenum]=(enode){head[v],u,0},head[v]=nodenum;
}
int bfs()
{for (int i=0;i<=N;i++) start[i]=head[i],dist[i]=INF;dist[S]=1,que.push(S);while (!que.empty()){int q=que.front(); que.pop();for (int i=head[q];i;i=e[i].nxt) {enode p=e[i];if (p.c<=0||dist[p.v]!=INF) continue;dist[p.v]=dist[q]+1;que.push(p.v);}}return dist[T]!=INF;
}
int dfs(int x,int T,int f)
{if (x==T) return f;int ans=0;for (int &i=start[x];i;i=e[i].nxt){if (e[i].c<=0||dist[e[i].v]!=dist[x]+1) continue;int t=dfs(e[i].v,T,min(f,e[i].c));e[i].c-=t,e[i^1].c+=t,f-=t,ans+=t;}return ans;
}
int dinic()
{int ans=0;while (bfs()) ans+=dfs(S,T,INF);return ans;
}
signed main()
{n=read(),R=read(),m=read(),S=1,T=2;for (int i=1;i<=5;i++) add(S,i+2,n/5);for (int i=1;i<=m;i++) a[i].fi=read(),a[i].se=read();a[++m]=MP(0,0),a[++m]=MP(R,n);sort(a+1,a+m+1);for (int j=2;j<=m;j++) {if (a[j].se<a[j-1].se) { puts("unfair"); return 0; }add(j+6,T,a[j].se-a[j-1].se);}for (int i=0;i<=4;i++) for (int j=2;j<=m;j++)add(i+3,j+6,get(a[j-1].fi+1,a[j].fi,i));N=m+6;if (dinic()==n) puts("fair");else puts("unfair");return 0;
}