Description
You have been given a matrix CNM, each element E of CNM is positive and no more than 1000, The problem is that if there exist N numbers a1, a2, … an and M numbers b1, b2, …, bm, which satisfies that each elements in row-i multiplied with ai and each elements in column-j divided by bj, after this operation every element in this matrix is between L and U, L indicates the lowerbound and U indicates the upperbound of these elements.
solution
**正解：差分约束
化简式子 (L<=a[i][j]*a[i]/b[j]<=U)
[a[i][j]*a[i]>=L*b[j]
]
[a[i][j]*a[i]<=U*b[j]
]
这样子并不好处理，差分约束式子一般表示为 (x+y<z) 的形式，所以想办法转化为加法
考虑取对数：
式子就变成了
[logai-logbj>=logL-loga[i][j]
]
[logai-logbj<=logU-loga[i][j]
]
ai到bj分别建边即可**
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#define RG register
#define il inline
using namespace std;
const int N=405,P=805,M=320005,inf=2e8;
int a[N][N],n,m,U,L,num=0,head[P],to[M],nxt[M],inst[P];
double dis[M];
void link(int x,int y,double z){
nxt[++num]=head[x];to[num]=y;dis[num]=z;head[x]=num;}
void Clear(){
memset(head,0,sizeof(head));
memset(inst,0,sizeof(inst));
num=0;
}
queue<int>q;
double f[P];int lim;bool vis[P];
bool spfa(){
while(!q.empty())q.pop();
q.push(1);
for(RG int i=1;i<=n+m;i++)
vis[i]=0,f[i]=inf;
RG int x,u;
f[1]=0;vis[1]=1;inst[1]++;
while(!q.empty()){
x=q.front();q.pop();
for(RG int i=head[x];i;i=nxt[i]){
u=to[i];
if(f[x]+dis[i]<f[u]){
f[u]=f[x]+dis[i];
if(!vis[u]){
vis[u]=1,q.push(u);
inst[u]++;
if(inst[u]>lim)return 0;
}
}
}
vis[x]=0;
}
return 1;
}
void work()
{
Clear();
double lu=log(U),lL=log(L);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++){
scanf("%d",&a[i][j]);
link(j+n,i,-lL+log(a[i][j]));
link(i,j+n,lu-log(a[i][j]));
}
lim=sqrt(n+m);
bool t=spfa();
if(t)puts("YES");
else puts("NO");
}
int main()
{
while(~scanf("%d%d%d%d",&n,&m,&L,&U))
work();
return 0;
}