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

Problem descriptiona 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.

 

Inputthe 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.

 

Outputfor 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.

 

Sample input3 4 0011 0001 1000 1100 4 0111 1001 1001 1110 3 000000 000

 

Sample outputyes No Yes HintIn 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. source2013 ACM-ICPC Jilin Tonghua national invitational competition-Questions reproduce to give you a picture, ask is not meet the requirements of the question, | A [I]-A [J] |> = T; for the edges in the vertex that are not in the vertex | A [I]-A [J] | <t | A [I] | <t; it is easy to know that if I-j has edges, a [I], a [J] symbols must be different, then this is a bipartite graph, the staining method can be used. If the dyeing method can be used, we can obtain a T value. Then construct the difference constraint equation and run the shortest short circuit to determine the negative circle.

#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 0x3f3f3f3fusing 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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HDU 4598 difference