HDU3666 差分约束

THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8418 Accepted Submission(s): 2179

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

题意：

n*m的矩阵c，是否存在n个数a1,a2,...an,和m个数b1,b2,...bm使得L<=cij*ai/bj<=R.

代码：

//显然不满足差分约束的条件，可以L<=cij*ai/bj<=R两边除cij（cij>0）后取对数得到
//log(L/cij)<=log(ai)-log(bj)<=log(R/cij).只求存不存在就行。但是本体如果用stl
//的queue写spfa会超时(可以用节点出队次数小于sqrt(n)判断)，可以自己定义一个栈来存储。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=402;
const double inf=100000008;
int n,m,tol,head[maxn*2+10],cnt[maxn*2+10],stk[maxn*maxn];
double L,R,dis[maxn*2+10];
bool mark[maxn*2+10];
struct node
{
    int to,next;
    double val;
}nodes[360010];
void Add(int a,int b,double c)
{
    nodes[tol].to=b;
    nodes[tol].val=c;
    nodes[tol].next=head[a];
    head[a]=tol++;
}
bool spfa(int s)
{
    for(int i=0;i<=n+m;i++){
        dis[i]=inf;
        cnt[i]=0;
        mark[i]=0;
    }
    int top=0;
    stk[++top]=s;
    mark[s]=1;cnt[s]++;dis[s]=0;
    while(top>0){
        int u=stk[top--];
        mark[u]=0;
        for(int i=head[u];i!=-1;i=nodes[i].next){
            int v=nodes[i].to;
            if(dis[v]>dis[u]+nodes[i].val){
                dis[v]=dis[u]+nodes[i].val;
                if(!mark[v]){
                    mark[v]=1;
                    if(++cnt[v]>=n+m) return 0;
                    stk[++top]=v;
                }
            }
        }
    }
    return 1;
}
int main()
{
    while(scanf("%d%d%lf%lf",&n,&m,&L,&R)==4){
        tol=0;
        memset(head,-1,sizeof(head));
        double x;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%lf",&x);
                Add(j+n,i,log(R/x));
                Add(i,j+n,-log(L/x));
            }
        }
        for(int i=1;i<=n+m;i++)//加一个公共源点0
            Add(0,i,0);
        if(spfa(0)) printf("YES
");
        else printf("NO
");
    }
    return 0;
}