HDU 3666 THE MATRIX PROBLEM （差分约束 深搜 & 广搜）

THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5437    Accepted Submission(s): 1372

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

 

题目意思就是是否存在ai,bj，使得l<=cij*(ai/bj)<=u (1<=i<=n,1<=j<=m)成立

首先，把cij除到两边:l'<=ai/bj<=u'，如果差分约束的话，应该是ai-bj的形式，于是可以取对数

log(l')<=log(ai)-log(bj)<=log(u')

把log(ai)和log(bj)看成两个点ai和bj，化成求最短路的形式：dis[ai]-dis[bj]<=log(u'),dis[bj]-dis[ai]<=-log(l')

然后判负环就行，深搜和广搜都可以

注意，如果spfa队列判负环：

(1)不必判断某个点入队次数大于N,只要判断是否大于sqrt(1.0*N)

(2)或者所有点的入队次数大于T*N，即存在负环，一般T取2

N为所有点的个数

 1, SPFA广搜：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>

using namespace std;

const int N=810;

struct Edge{
    int to,nxt;
    double cap;
}edge[N*N];

int n,m,cnt,head[N];
int vis[N],Count[N];
double dis[N],L,U;

void addedge(int cu,int cv,double cw){
    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
}

int SPFA(){
    int limit=(int)sqrt(1.0*(n+m));
    queue<int> q;
    while(!q.empty())
        q.pop();
    memset(vis,0,sizeof(vis));
    memset(Count,0,sizeof(Count));
    for(int i=0;i<=n+m;i++){
        dis[i]=0;
        q.push(i);
    }
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].to;
            if(dis[v]>dis[u]+edge[i].cap){
                dis[v]=dis[u]+edge[i].cap;
                if(!vis[v]){
                    vis[v]=1;
                    if(++Count[v]>limit)
                        return 0;
                    q.push(v);
                }
            }
        }
    }
    return 1;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d%lf%lf",&n,&m,&L,&U)){
        cnt=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);
                addedge(j+n,i,log(U/x));
                addedge(i,j+n,-log(L/x));
            }
        if(SPFA())
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

2, SPFA深搜：（这个更快？？）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>

using namespace std;

const int N=810;

struct Edge{
    int to,nxt;
    double cap;
}edge[N*N];

int n,m,cnt,head[N];
int vis[N],instack[N];
double dis[N],L,U;

void addedge(int cu,int cv,double cw){
    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
}

int SPFA(int u){
    if(instack[u])
        return 0;
    instack[u]=1;
    vis[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        if(dis[v]>dis[u]+edge[i].cap){
            dis[v]=dis[u]+edge[i].cap;
            if(!SPFA(v))
                return 0;
        }
    }
    instack[u]=0;
    return 1;
}

int solve(){
    memset(vis,0,sizeof(vis));
    memset(instack,0,sizeof(instack));
    memset(dis,0,sizeof(dis));
    for(int i=1;i<=n+m;i++)
        if(!vis[i]){
            if(!SPFA(i))
                return 0;
        }
    return 1;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d%lf%lf",&n,&m,&L,&U)){
        cnt=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);
                addedge(j+n,i,log(U/x));
                addedge(i,j+n,-log(L/x));
            }
        if(solve())
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}