THE MATRIX PROBLEM
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8016 Accepted Submission(s): 2092
Problem Description
You have been given a matrix CN*M, each element E of CN*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
题意:给出一个矩阵 C 问是否存在这样两个序列使得 l <= cij * ai/bj <= r 对这个矩阵成立。
题解:将cij除过去,然后取个对数， log(l/cij) <= log(ai) - log(bj) <= log(r/cij) 这样的话就可以化为一个差分约束的题了,但是这个题有个问题就是判环会超时,有人提出可以只判断sqrt(n)次就行了，，但是无法证明,这个题的更好的方法是用一个手动栈代替队列.这样的话就能够快速判环了，给出两份代码..
只用sqrt(n)的队列写法:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<math.h>
using namespace std;
const double INF = 999999999;
const int N = 805;
const int M = 800000;
struct Edge{
int v,next;
double w;
}edge[M];
int head[N],tot;
int n,m,l,r;
void init(){
memset(head,-1,sizeof(head));
tot = 0;
}
void addEdge(int u,int v,double w,int &k){
edge[k].v =v ,edge[k].w = w ,edge[k].next = head[u],head[u] = k++;
}
double low[N];
int time[N];
bool vis[N];
bool spfa(int s){
for(int i=0;i<=n+m;i++){
vis[i] = false;
time[i] = 0;
low[i] = INF;
}
int num = ((int)sqrt(n+m))+1;
low[s] = 0;
time[s]++;
queue<int>q;
q.push(s);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
for(int k=head[u];k!=-1;k=edge[k].next){
int v = edge[k].v;
double w = edge[k].w;
if(low[v]>low[u]+w){
low[v] = low[u]+w;
if(!vis[v]){
vis[v] = true;
q.push(v);
time[v]++;
if(time[v]>num) return false;
}
}
}
}
return true;
}
int main()
{
double c;
while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF){
init();
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%lf",&c);
addEdge(i,n+j,-log(l/c),tot);
addEdge(n+j,i,log(r/c),tot);
}
}
for(int i=1;i<=n+m;i++){
addEdge(0,i,0,tot);
}
if(spfa(0)) printf("YES\n");
else printf("NO\n");
}
return 0;
}
手动栈解决:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<math.h>
using namespace std;
const double INF = 999999999;
const int N = 805;
const int M = 800000;
struct Edge{
int v,next;
double w;
}edge[M];
int head[N],tot;
int n,m,l,r;
void init(){
memset(head,-1,sizeof(head));
tot = 0;
}
void addEdge(int u,int v,double w,int &k){
edge[k].v =v ,edge[k].w = w ,edge[k].next = head[u],head[u] = k++;
}
double low[N];
int time[N];
bool vis[N];
int stk[N*N];
bool spfa(int s){
for(int i=0;i<=n+m;i++){
vis[i] = false;
time[i] = 0;
low[i] = INF;
}
int top = 0;
low[s] = 0;
time[s]++;
stk[top++] = s;
while(top!=0){
int u = stk[--top];
vis[u] = false;
for(int k=head[u];k!=-1;k=edge[k].next){
int v = edge[k].v;
double w = edge[k].w;
if(low[v]>low[u]+w){
low[v] = low[u]+w;
if(!vis[v]){
vis[v] = true;
stk[top++] = v;
time[v]++;
if(time[v]>n+m) return false;
}
}
}
}
return true;
}
int main()
{
double c;
while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF){
init();
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%lf",&c);
addEdge(i,n+j,-log(l/c),tot);
addEdge(n+j,i,log(r/c),tot);
}
}
for(int i=1;i<=n+m;i++){
addEdge(0,i,0,tot);
}
if(spfa(0)) printf("YES\n");
else printf("NO\n");
}
return 0;
}