HDOJ 3666 THE MATRIX PROBLEM 差分约束

根据题意有乘除的关系,为了方便构图,用对数转化乘除关系为加减关系.....

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 CN*M, each element E of CN*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

3 3 1 6
2 3 4
8 2 6
5 2 9

 

Sample Output

YES

 

Source

2010 Asia Regional Harbin

 

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

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<n+m+10;i++)
dist[i]=1e40;
memset(cQ,0,sizeof(cQ));
memset(inQ,false,sizeof(inQ));

dist[rt]=0;
queue<int> 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;
}