HDU 3666 THE MATRIX PROBLEM 差分约束系统

 

THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3001 Accepted Submission(s): 784

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

   
   
   
   

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

Sample Output

   
   
   
   

    
    
    
    YES
 
   
   
   
   

Source

2010 Asia Regional Harbin

Recommend

lcy

 

给一个n*m的矩阵X和数L、U，求n个数ai，和m个数bj，使得Xij*ai/bj在L到U的范围内。

刚开始看到不等式的时候就猜测可能是差分约束系统，但是不等式里面有乘法不知道怎么搞，后来才知道通过取对数把乘法转换为加法就行了。

下面来看不等式：

L<=Xij*ai/bj<=U              ==>

L*bj<=Xij*ai<=U*bj         ==>

L*bj-Xij*ai<=0

Xij*ai-U*bj<=0                 ==>

log(L*bj)-log(Xij*ai)<=0

log(Xij*ai)-log(U*bj)<=0   ==>

logbj-logai<=logXij-logL

logai-logbj<=logU-logXij

这样就转换为标准的差分约束系统的不等式。

所以建图，有n+m个节点

i到j+n连边，边权为log(Xij/L)

j+n到i连边，边权为log(U/Xij)

然后用spfa判断是否有负环，有负环说明不等式组无解。

但是普通的bellman，spfa判负环都会超时，递归的spfa也不行，

只有使用经验公式了：

1.cnt[]记录节点加入队列次数，大于sqrt(n)求可以当做有负环了。

2.sum记录所有节点加入队列次数大于2*(n+m),也当做有负环了。

但这是不正确的，要看数据了。

 

代码：

开了输入外挂才优化到400ms...

#include<cstdio>
#include<cstring>
#include<cmath>
#define inf 999999999
using namespace std;

int n,m,num,adj[805],f[805],q[10005];
double dis[805],L,U;
struct edge
{
	int v,pre;
	double w;
}e[320005];
void insert(int u,int v,double w)
{
	e[num].v=v;
	e[num].w=w;
	e[num].pre=adj[u];
	adj[u]=num++;
}
int spfa(int x)
{
	int i,v,sum=0,head=0,tail=0;
	memset(f,0,sizeof(f));
	for(i=0;i<n;i++)
		dis[i]=inf;
	dis[0]=0;
	q[tail++]=x;
	while(head<tail)
	{
		x=q[head++];
		f[x]=0;
		for(i=adj[x];~i;i=e[i].pre)
			if(dis[v=e[i].v]>dis[x]+e[i].w+1e-8)
			{
				dis[v]=dis[x]+e[i].w;
				if(!f[v])
				{
					sum++;
					if(sum>2*(n+m))
						return 0;
					f[v]=1;
					q[tail++]=v;
				}
			}
	}
	return 1;
}
inline int getint()
{
    char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    int n=c-'0';
    c=getchar();
    while(c>='0'&&c<='9')
    {
        n=(n<<1)+(n<<3)+c-'0';
        c=getchar();
    }
    return n;
}
int main()
{
	while(~scanf("%d",&n))
	{
		int i,j,k;
		num=0;
		memset(adj,-1,sizeof(adj));
		m=getint();
		L=getint();
		U=getint();
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				k=getint();
				insert(i,j+n,log(k/L));
				insert(j+n,i,log(U/k));
			}
		n+=m;
		m+=m;
		if(spfa(0))
			puts("YES");
		else
			puts("NO");
	}
}