Problem Description

You have been given a matrix C_N*M, each element E of C_N*M 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.

 

Input

There are several test cases. You should process to the end of file.
Each case includes two parts, in part 1, there are four integers in one line, N,M,L,U, indicating the matrix has N rows and M columns, L is the lowerbound and U is the upperbound (1<=N、M<=400,1<=L<=U<=10000). In part 2, there are N lines, each line includes M integers, and they are the elements of the matrix.

 

Output

If there is a solution print "YES", else print "NO".

 

Sample Input

 

Sample Output

 

Source

 

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#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<cmath>

using namespace std;

const int N=;

struct Edge{

    int to,nxt;

    double cap;

}edge[N*N];

int n,m,cnt,head[N];

int vis[N],Count[N];

double dis[N],L,U;

void addedge(int cu,int cv,double cw){

    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];

    head[cu]=cnt++;

}

int SPFA(){

    int limit=(int)sqrt(1.0*(n+m));

    queue<int> q;

    while(!q.empty())

        q.pop();

    memset(vis,,sizeof(vis));

    memset(Count,,sizeof(Count));

    for(int i=;i<=n+m;i++){

        dis[i]=;

        q.push(i);

    }

    while(!q.empty()){

        int u=q.front();

        q.pop();

        vis[u]=;

        for(int i=head[u];i!=-;i=edge[i].nxt){

            int v=edge[i].to;

            if(dis[v]>dis[u]+edge[i].cap){

                dis[v]=dis[u]+edge[i].cap;

                if(!vis[v]){

                    vis[v]=;

                    if(++Count[v]>limit)

                        return ;

                    q.push(v);

                }

            }

        }

    }

    return ;

}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d%lf%lf",&n,&m,&L,&U)){

        cnt=;

        memset(head,-,sizeof(head));

        double x;

        for(int i=;i<=n;i++)

            for(int j=;j<=m;j++){

                scanf("%lf",&x);

                addedge(j+n,i,log(U/x));

                addedge(i,j+n,-log(L/x));

            }

        if(SPFA())

            puts("YES");

        else

            puts("NO");

    }

    return ;

}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<cmath>

using namespace std;

const int N=;

struct Edge{

    int to,nxt;

    double cap;

}edge[N*N];

int n,m,cnt,head[N];

int vis[N],instack[N];

double dis[N],L,U;

void addedge(int cu,int cv,double cw){

    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];

    head[cu]=cnt++;

}

int SPFA(int u){

    if(instack[u])

        return ;

    instack[u]=;

    vis[u]=;

    for(int i=head[u];i!=-;i=edge[i].nxt){

        int v=edge[i].to;

        if(dis[v]>dis[u]+edge[i].cap){

            dis[v]=dis[u]+edge[i].cap;

            if(!SPFA(v))

                return ;

        }

    }

    instack[u]=;

    return ;

}

int solve(){

    memset(vis,,sizeof(vis));

    memset(instack,,sizeof(instack));

    memset(dis,,sizeof(dis));

    for(int i=;i<=n+m;i++)

        if(!vis[i]){

            if(!SPFA(i))

                return ;

        }

    return ;

}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d%lf%lf",&n,&m,&L,&U)){

        cnt=;

        memset(head,-,sizeof(head));

        double x;

        for(int i=;i<=n;i++)

            for(int j=;j<=m;j++){

                scanf("%lf",&x);

                addedge(j+n,i,log(U/x));

                addedge(i,j+n,-log(L/x));

            }

        if(solve())

            puts("YES");

        else

            puts("NO");

    }

    return ;

}