hdu 4598 Difference

Difference

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 621 Accepted Submission(s): 161


[align=left]Problem Description[/align]
A graph is a difference if every vertex vi can be assigned a real number ai and there exists a positive real number T such that
(a) |ai| < T for all i and
(b) (vi, vj) in E <=> |ai - aj| >= T,
where E is the set of the edges.
Now given a graph, please recognize it whether it is a difference.

[align=left]Input[/align]
The first line of input contains one integer TC(1<=TC<=25), the number of test cases.
Then TC test cases follow. For each test case, the first line contains one integer N(1<=N<=300), the number of vertexes in the graph. Then N lines follow, each of the N line contains a string of length N. The j-th character in the i-th line is "1" if (vi, vj) in E, and it is "0" otherwise. The i-th character in the i-th line will be always "0". It is guaranteed that the j-th character in the i-th line will be the same as the i-th character in the j-th line.

[align=left]Output[/align]
For each test case, output a string in one line. Output "Yes" if the graph is a difference, and "No" if it is not a difference.

[align=left]Sample Input[/align]

3
4
0011
0001
1000
1100

4
0111
1001
1001
1110

3
000

000
000

[align=left]Sample Output[/align]

Yes
No
Yes

Hint

In sample 1, it can let T=3 and a[sub]1[/sub]=-2, a[sub]2[/sub]=-1, a[sub]3[/sub]=1, a[sub]4[/sub]=2.

[align=left]Source[/align]
2013 ACM-ICPC吉林通化全国邀请赛——题目重现

给你一个图，问是不是满足题目要求的条件，
|a[i]-a[j]| >= T ;对于在图里的边，不在图里的边|a[i]-a[j]| < T
| a[i] | < T ;
我们很容易知道如果 i-j有边，a[i],a[j]符号肯定不同，那么这个就是二分图了，染色法可以做
如果可以染色 我们可以随便取个 T 值。然后构造差分约束方程，
跑最短路判负圈

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define mod 1000000007
#define maxn 410
#define pi acos(-1.0)
#define eps 1e-8
#define INF 0x3f3f3f3f
using namespace std;

vector<int>qe[maxn] ;
int color[maxn] ,flag,top,val[maxn*maxn] ;
int head[maxn],to[maxn*maxn],next1[maxn*maxn];
char a[maxn][maxn] ;

void Unit(int u,int v,int c)
{
next1[top] = head[u] ;to[top] = v ;
val[top]=c;head[u]=top++;
}
void dfs(int u,int fa,int c )
{
color[u]=c;
for(int i = 0 ; i < qe[u].size();i++)
{
int v = qe[u][i] ;
if(v==fa) continue ;
if(color[v]==-1)
{
dfs(v,u,1-c) ;
}
else if(color[v]==c)
{
flag=1;
return ;
}
}
}
int dis[maxn] ,cnt[maxn];
bool vi[maxn];
bool spfa(int s,int n)
{
queue<int>q;
for(int i = 1 ; i <= n ;i++)
{
dis[i]=0;
vi[i]=1;
q.push(i);
cnt[i]=0;
}
int i,u,v;
while(!q.empty())
{
u  = q.front();q.pop();
for( i = head[u] ; i != -1 ; i = next1[i])
{
v = to[i] ;
if(dis[v] > dis[u]+val[i])
{
dis[v]=dis[u]+val[i] ;
if(!vi[v])
{
vi[v]=true;
cnt[v]++;
if(cnt[v]>n) return false;
q.push(v) ;
}
}
}
vi[u]=false;
}
return true;
}
int main()
{
int j,i,l,g,n;
int T,ans1,u,v;
cin >> T ;
while(T--)
{
scanf("%d",&n) ;
for( i = 1 ; i <= n ;i++){
scanf("%s",a[i]+1) ;
qe[i].clear();
}
for( i = 1 ; i <= n ;i++)
for( j = 1 ; j <= n ;j++)if(a[i][j]=='1'){
qe[i].push_back(j);
}
memset(color,-1,sizeof(color)) ;
flag=0;
for( i = 1 ; i <= n ;i++)if(color[i]==-1){
dfs(i,-1,1) ;
if(flag) break ;
}
if(flag){
puts("No") ;
continue ;
}
top=0;
memset(head,-1,sizeof(head)) ;
for( i = 1 ; i <= n ;i++)
for( j = i+1 ; j <= n ;j++)
{
if(a[i][j]=='1')
{
if(color[i])
Unit(i,j,-maxn) ;
else Unit(j,i,-maxn) ;
}
else
{
if(color[i])
Unit(j,i,maxn-1);
else Unit(i,j,maxn-1) ;
}
}
if(spfa(1,n))puts("Yes") ;
else puts("No") ;
}
return 0 ;
}

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