G - THE MATRIX PROBLEM(差分约束&判负环优化)

然后转差分约束判负环即可，最好建立一个超级源点，避免图不连通。

本题需要判负环的时候优化，即： + + i n [ v ] > s q r t ( n + m ) ++in[v]>sqrt(n+m) ++in[v]>sqrt(n+m) 返回 f a l s e false false。

// Problem: G - THE MATRIX PROBLEM
// Contest: Virtual Judge - 2010icpc [Cloned]
// URL: https://vjudge.net/contest/467412#problem/G
// Memory Limit: 32 MB
// Time Limit: 2000 ms
// Date: 2021-11-07 20:11:34
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e3+5,M=2e6+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = {402653189,805306457,1610612741,998244353};
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n){
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\n",a[n]); 
}
#define db double
db lgL,lgR;
int h[N],cnt,n,m,L,R;
db d[N];
struct edge{
	int to,nt;
	db w;
}e[M<<1];
void add(int u,int v,db w){
	e[++cnt]={v,h[u],w},h[u]=cnt;
}
bitset<N>vis;
int in[N];
void bud(){
	for(int i=1;i<=n+m;i++) add(0,i,0);
	rep(i,1,n){
		rep(j,1,m){
			int x;scanf("%d",&x);
			db lgx=log(1.0*x);
			add(n+j,i,lgR-lgx);
			add(i,n+j,lgx-lgL);
		}
	}
}
int sqk;
bool spfa(int st,int ed){
	mst(in,0);vis.reset();
	queue<int>q;
	rep(i,1,n+m) d[i]=inf;
	d[0]=0; 
	vis[st]=1;q.push(st);
	while(!q.empty()){
		int u=q.front();q.pop();vis[u]=0;
		for(int i=h[u];i;i=e[i].nt){
			int v=e[i].to;
			db w=e[i].w;
			if(d[v]>d[u]+w){
				d[v]=d[u]+w;
				if(!vis[v]) {
					if(++in[v]>sqk) return false;
					q.push(v),vis[v]=1;
				}
			}
		}
	}
	return true;
}
int main(){
	while(~scanf("%d%d%d%d",&n,&m,&L,&R)){
		cnt=0;mst(h,0);
		lgL=log(L),lgR=log(R);
		sqk=sqrt(n+m);
		bud();
		puts(spfa(0,n+m)?"YES":"NO");
	}
	return 0;
}