HDU 3666 THE MATRIX PROBLEM（差分约束）

题意：给一个 n*m 的矩阵，问是否能够给每行乘以一个数 Xi，给每列除以一个数 Yj（两行两列之间可以不同），使得最后矩阵中所有元素的值在区间 [L,U] 内

思路：这里的话有除法不好处理，所以两边直接取对数，转化为d[ai]>=d[bi]+log(l/cij),d[bj]>=d[ai]-log(r/cij)这样求最长路或者求最短路也是可以的，代码里面用的是求最短路的方法，然后只是找环的话用DFS会比BFS快很多，直接BFS的话会超时

#include<bits/stdc++.h>
using namespace std;
const int maxn = 805;
vector<pair<int,double> >e[maxn];
/*struct Edge
{
int v,net;
double w;
}edge[maxn*maxn*2];
int cnt,head[maxn];*/
double d[maxn];
int inq[maxn],n,m;
int num[maxn];
/*void init()
{
cnt = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,double w)
{
edge[cnt].v=v,edge[cnt].w=w;
edge[cnt].net=head[u];
head[u]=cnt++;
}*/

bool dfs(int u)
{
//if(inq[u])return true;
inq[u]=1;
for(int i = 0;i<e[u].size();i++)
// for(int i = head[u];i!=-1;i=edge[i].net)
{
int v = e[u][i].first;
// int v = edge[i].v;
// double w = edge[i].w;
if(d[v]>d[u]+e[u][i].second)
// if(d[v]>d[u]+w)
{
d[v]=d[u]+e[u][i].second;
// if(++num[v]>n+m)return true;
//d[v]=d[u]+e[u][i].second;
if(inq[v] || dfs(v))
return true;
}
}
inq[u]=0;
return false;
}
bool spfa()
{
memset(inq,0,sizeof(inq));
for(int i = 0;i<=n+m;i++)
d[i]=1e8;
d[0]=0;
//memset(d,0,sizeof(d));
for(int i = 0;i<=n+m;i++)
if(dfs(i))
return true;
return false;
}
int main()
{
int l,r;
while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF)
{
//init();
for(int i = 0;i<=n+m+1;i++)
e[i].clear();
for(int i = 1;i<=n;i++)
for(int j = 1;j<=m;j++)
{
int x;
scanf("%d",&x);
// addedge(i,j+n,-log(1.0*l/x));
// addedge(j+n,i,log(1.0*r/x));
e[i].push_back(make_pair(j+n,-log(1.0*l/x)));
e[j+n].push_back(make_pair(i,log(1.0*r/x)));
}
for(int i = 1;i<=n+m;i++)
// addedge(i,0,0);
e[i].push_back(make_pair(0,0));
if(spfa())
printf("NO\n");
else
printf("YES\n");
}
}


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