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HDOJ 3666 THE MATRIX PROBLEM 差分約束

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HDOJ 3666 THE MATRIX PROBLEM 差分約束

根據題意有乘除的關系,為了方便構圖,用對數轉化乘除關系為加減關系.....

THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7486 Accepted Submission(s): 1914

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

YES

Source 2010 Asia Regional Harbin

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月29日 星期三 20時55分16秒
File Name     :HDOJ3666.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

using namespace std;

const int maxn=1000;
const double eps=1e-8;

int n,m;
double L,R;

struct Edge
{
	int to,next;
	double cost;
}edge[maxn*maxn*2];

int Adj[maxn],Size;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void Add_Edge(int u,int v,double c)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].cost=c;
	Adj[u]=Size++;
}

double dist[maxn];
int cQ[maxn];
bool inQ[maxn];

bool spfa(int rt)
{	
	for(int i=0;i q;
    inQ[rt]=true;q.push(rt); cQ[rt]=1;

    while(!q.empty())
    {
        int u=q.front();q.pop();

        for(int i=Adj[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dist[v]>dist[u]+edge[i].cost)
            {
                dist[v]=dist[u]+edge[i].cost;
                if(!inQ[v])
                {
                    inQ[v]=true;
                    cQ[v]++;
                    if(cQ[v]>=sqrt(n+m)) return false;
                    q.push(v);
                }
            }
        }
        inQ[u]=false;
    }
    return true;
}

int main()
{
	//freopen(in.txt,r,stdin);
	//freopen(out.txt,w,stdout);

	while(scanf(%d%d%lf%lf,&n,&m,&L,&R)!=EOF)
	{
		init();

		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				double x;
				scanf(%lf,&x);
				int a=i,b=n+j;
				Add_Edge(b,a,log(R/x));
				Add_Edge(a,b,-log(L/x));
			}
		}

		for(int i=1;i<=n+m;i++) 
			Add_Edge(0,i,0);

		bool fg=spfa(0);

		if(fg) puts(YES);
		else puts(NO);
	}
    
    return 0;
}