hdoj--3666--THE MATRIX PROBLEM(差分约束+SPFA深搜)
THE MATRIX PROBLEM
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.
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.
3 3 1 6
2 3 4
8 2 6
5 2 9
YES
超时代码:(应该是oj编译器问题或者就是AC的标准提高了)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAXN 1000
#define MAXM 500000+10
#define INF 0x3f3f3f
int dis[MAXN],vis[MAXN],used[MAXN],m,n;
int head[MAXN],cnt;
double map[MAXN][MAXN];
double L,U;
struct node
{
int u,v;
double val;
int next;
}edge[MAXM];
void init()
{
memset(head,-1,sizeof(head));
memset(map,0,sizeof(map));
cnt=0;
}
void add(int u,int v,int val)
{
node E={u,v,val,head[u]};
edge[cnt]=E;
head[u]=cnt++;
}
void getmap()
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%lf",&map[i][j]);
add(j+n,i,log(U/map[i][j]));
add(i,j+n,-log(L/map[i][j]));
}
}
for(int i=1;i<=n+m;i++)
add(0,i,0);
}
void SPFA()
{
memset(vis,0,sizeof(vis));
memset(used,0,sizeof(used));
memset(dis,INF,sizeof(dis));
queue<int>q;
q.push(0);
dis[0]=0;
used[0]++;
vis[0]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
node E=edge[i];
if(dis[E.v]>dis[E.u]+E.val)
{
dis[E.v]=dis[E.u]+E.val;
if(!vis[E.v])
{
vis[E.v]=1;
used[E.v]++;
if(used[E.v]>(int)sqrt(1.0*n+m))
{
cout<<"NO"<<endl;
return ;
}
q.push(E.v);
}
}
}
}
cout<<"YES"<<endl;
}
int main()
{
while(scanf("%d%d%lf%lf",&n,&m,&L,&U)!=EOF)
{
init();
getmap();
SPFA();
}
return 0;
}
SPFA深搜版
#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stack>
#include<queue>
using namespace std;
const int MAX=805;
struct node
{
int v,next;
double c;
}g[MAX*MAX];
int adj[MAX];
int n,m,e;
double dis[MAX],l,u;
bool vis[MAX],inStack[MAX];
inline void add(int u,int v,double c)
{
g[e].v=v; g[e].c=c; g[e].next=adj[u]; adj[u]=e++;
}
bool spfa(int u)
{
int i,v;
if(inStack[u])
return false;
inStack[u]=true;
vis[u]=true;
for(i=adj[u];i!=-1;i=g[i].next)
{
v=g[i].v;
if(dis[v]>dis[u]+g[i].c)
{
dis[v]=dis[u]+g[i].c;
if(!spfa(v))
{
return false;
}
}
}
inStack[u]=false;
return true;
}
bool ok()
{
int i,u,v,cnt=0;
memset(vis,0,sizeof(vis));
memset(inStack,0,sizeof(inStack));
for(i=0;i<=n+m;i++)
{
dis[i]=0;
}
for(i=1;i<=n+m;i++)
{
if(!vis[i])
{
if(!spfa(i))
{
return false;
}
}
}
return true;
}
int main()
{
int i,j;
double t;
while(scanf("%d%d %lf %lf",&n,&m,&l,&u)!=EOF)
{
e=0;
memset(adj,-1,sizeof(adj));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%lf",&t);
add(j+n,i,log(u/t));
add(i,j+n,-log(l/t));
}
}
if(ok())
puts("YES");
else
puts("NO");
}
return 0;
}
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