HDU 4751 2-sat模板题

Divide Groups

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 161    Accepted Submission(s): 62

Problem Description

  This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos.   After carefully planning, Tom200 announced his activity plan, one that contains two characters:   1. Whether the effect of the event are good or bad has nothing to do with the number of people join in.   2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other.   The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one's energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large.   Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.

 

 

Input

  The input contains several test cases, terminated by EOF.   Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event.   N lines follow. The i-th line contains some integers which are the id of students that the i-th student knows, terminated by 0. And the id starts from 1.

 

 

Output

  If divided successfully, please output "YES" in a line, else output "NO".

 

 

Sample Input

3 3 0 1 0 1 2 0

 

 

Sample Output

YES

由于数据范围很小，直接开二维数组标记两个节点之间的关系，由于题目要求一个组内任意两个人都相互认识，因此对于任意两人，只要不是相互认识，就根据矛盾关系建边，然后2-sat判定。

贴两份代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<string>
using namespace std;
int head[250],tol,mark[2000],S[2000],c,n,flag[300][300];
struct node
{
        int next,to;
}edge[900000];
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}

bool dfs(int x)
{
        if(mark[x^1])return 0;
        if(mark[x])return 1;
        mark[x]=1;
        S[c++]=x;
        for(int i=head[x];i!=-1;i=edge[i].next)
        {
                int v=edge[i].to;
                if(!dfs(v))return 0;
        }
        return 1;
}

bool solve()
{
        for(int i=0;i<2*n;i+=2)
        if(!mark[i]&&!mark[i+1])
        {
                c=0;
                if(!dfs(i))
                {
                        while(c>0)mark[S[--c]]=0;
                        if(!dfs(i+1))return 0;
                }
        }
        return 1;
}

int main()
{
        int i,j,k,T,t;
        while(~scanf("%d",&n))
        {
                memset(flag,0,sizeof(flag));
                memset(head,-1,sizeof(head));tol=0;
                memset(mark,0,sizeof(mark));c=0;
                for(i=1;i<=n;i++)
                {
                       while(~scanf("%d",&j)&&j)flag[i][j]=1;
                }
                for(i=1;i<=n;i++)
                for(j=i+1;j<=n;j++)
                if(flag[i][j]==0||flag[j][i]==0)
                {
                        int p,q;
                        p=i-1;q=j-1;
                        add(2*p,2*q+1);
                        add(2*q,2*p+1);
                        add(2*p+1,2*q);
                        add(2*q+1,2*p);
                }
                if(solve())puts("YES");
                else puts("NO");
        }
        return 0;
}

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<string>
using namespace std;
struct node
{
        int to,next;
}edge[900000];
int head[300],tol,low[300],dfn[300],indexx,Stack[300],instack[300],belong[300],scc,top,flag[300][300],n;
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}
void tarjin(int u)
{
        low[u]=dfn[u]=++indexx;
        Stack[top++]=u;instack[u]=1;
        int i,v;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
                v=edge[i].to;
                if(!dfn[v])
                {
                        tarjin(v);
                        if(low[u]>low[v])low[u]=low[v];
                }
                else if(instack[v]&&low[u]>dfn[v])low[u]=dfn[v];
        }
        if(low[u]==dfn[u])
        {
                scc++;
                do
                {
                        v=Stack[--top];
                        instack[v]=0;
                        belong[v]=scc;
                }while(u!=v);
        }
}
int solve()
{
        memset(dfn,0,sizeof(dfn));
        memset(instack,0,sizeof(instack));
        memset(belong,0,sizeof(belong));
        indexx=top=scc=0;
        for(int i=0;i<2*n;i++)if(!dfn[i])tarjin(i);
        //cout<<tol<<" "<<scc<<endl;
        for(int i=0;i<n;i++)if(belong[2*i]==belong[2*i+1])return 0;
        return 1;
}
int main()
{
        int i,j,k,T,t;
        while(~scanf("%d",&n))
        {
                memset(flag,0,sizeof(flag));
                memset(head,-1,sizeof(head));tol=0;
                for(i=1;i<=n;i++)
                {
                       while(~scanf("%d",&j)&&j)flag[i][j]=1;
                }
                for(i=1;i<=n;i++)
                for(j=i+1;j<=n;j++)
                if(flag[i][j]==0||flag[j][i]==0)
                {
                        int p,q;
                        p=i-1;q=j-1;
                        add(2*p,2*q+1);
                        add(2*q,2*p+1);
                        add(2*p+1,2*q);
                        add(2*q+1,2*p);
                }
                if(solve())puts("YES");
                else puts("NO");
        }
        return 0;
}