hdu 4115 Eliminate the Conflict （ 2-sat ）

Eliminate the Conflict

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1315    Accepted Submission(s): 563

Problem Description

Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated.
Edward contributes his lifetime to invent a 'Conflict Resolution Terminal' and he has finally succeeded. This magic item has the ability to eliminate all the conflicts. It works like this:
If any two people have conflict, they should simply put their hands into the 'Conflict Resolution Terminal' (which is simply a plastic tube). Then they play 'Rock, Paper and Scissors' in it. After they have decided what they will play, the tube should be opened and no one will have the chance to change. Finally, the winner have the right to rule and the loser should obey it. Conflict Eliminated!
But the game is not that fair, because people may be following some patterns when they play, and if the pattern is founded by others, the others will win definitely.
Alice and Bob always have conflicts with each other so they use the 'Conflict Resolution Terminal' a lot. Sadly for Bob, Alice found his pattern and can predict how Bob plays precisely. She is very kind that doesn't want to take advantage of that. So she tells Bob about it and they come up with a new way of eliminate the conflict:
They will play the 'Rock, Paper and Scissors' for N round. Bob will set up some restricts on Alice.
But the restrict can only be in the form of "you must play the same (or different) on the ith and jth rounds". If Alice loses in any round or break any of the rules she loses, otherwise she wins.
Will Alice have a chance to win?

 

Input

The first line contains an integer T(1 <= T <= 50), indicating the number of test cases.
Each test case contains several lines.
The first line contains two integers N,M(1 <= N <= 10000, 1 <= M <= 10000), representing how many round they will play and how many restricts are there for Alice.
The next line contains N integers B ₁,B ₂, ...,B _N, where B _i represents what item Bob will play in the i ^th round. 1 represents Rock, 2 represents Paper, 3 represents Scissors.
The following M lines each contains three integers A,B,K(1 <= A,B <= N,K = 0 or 1) represent a restrict for Alice. If K equals 0, Alice must play the same on A ^th and B ^th round. If K equals 1, she must play different items on Ath and Bthround.

 

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y is "yes" or "no" represents whether Alice has a chance to win.

 

Sample Input

   
   
   
   

    
    
    
    2
3 3
1 1 1
1 2 1
1 3 1
2 3 1
5 5
1 2 3 2 1
1 2 1
1 3 1
1 4 1
1 5 1
2 3 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: no
Case #2: yes

    
    
    
    
 
     
 
      Hint 
     
 'Rock, Paper and Scissors' is a game which played by two person. They should play Rock, Paper or Scissors by their hands at the same time. Rock defeats scissors, scissors defeats paper and paper defeats rock. If two people play the same item, the game is tied.. 
    
    
    
    
     
   
   
   
   

 

Source

2011 Asia ChengDu Regional Contest

题意：

Alice和Bob玩石头、剪刀、布，玩n局，Alice已知Bob的n局的出法，Bob给了Alice m个限制第i、j两次必须相同或不同，问Alice能否不输。

思路：

已知Bob的出法后，保证Alice不输，那每局Alice就有两种出法，问题就变为2-sat了，根据Alice每局的出法，给出限制后找到约束条件建边就够了。注意挖掘隐含的约束条件，如i、j相同，而Bob在i、j出法不同，那么Alice在i中的选择与在j中的选择只有一个交集，必须选择这个交集，选了其他的就错了。（我就卡在这个地方了）

代码：

#include <cstdio>
#include <cstring>
#define maxn 20005
#define MAXN 100005
using namespace std;

int n,m,num,flag;
int bob[maxn];
int head[maxn];
int scc[maxn];
int vis[maxn];
int stack1[maxn];
int stack2[maxn];
struct edge
{
    int v,next;
} g[MAXN];
int a[4][2]=
{
    0,0,
    1,2,
    2,3,
    1,3
};

void init()
{
    memset(head,0,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(scc,0,sizeof(scc));
    stack1[0] = stack2[0] = num = 0;
    flag = 1;
}
void addedge(int u,int v)
{
    num++;
    g[num].v = v;
    g[num].next = head[u];
    head[u] = num;
}
void dfs(int cur,int &sig,int &cnt)
{
    if(!flag) return;
    vis[cur] = ++sig;
    stack1[++stack1[0]] = cur;
    stack2[++stack2[0]] = cur;
    for(int i = head[cur]; i; i = g[i].next)
    {
        if(!vis[g[i].v]) dfs(g[i].v,sig,cnt);
        else
        {
            if(!scc[g[i].v])
            {
                while(vis[stack2[stack2[0]]] > vis[g[i].v])
                    stack2[0] --;
            }
        }
    }
    if(stack2[stack2[0]] == cur)
    {
        stack2[0] --;
        ++cnt;
        do
        {
            scc[stack1[stack1[0]]] = cnt;
            int tmp = stack1[stack1[0]];
            if((tmp >n && scc[tmp - n] == cnt) || (tmp <= n && scc[tmp + n] == cnt))  // 注意这里的‘=’号
            {
                flag = false;
                return;
            }
        }
        while(stack1[stack1[0] --] != cur);
    }
}
void Twosat()
{
    int i,sig,cnt;
    sig = cnt = 0;
    for(i=0; i<n+n&&flag; i++)
    {
        if(!vis[i]) dfs(i,sig,cnt);
    }
}
int main()
{
    int i,j,t,u,v,k,test=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        init();
        num=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&bob[i]);
        }
        for(i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&k);
            if(k)
            {
                if(a[bob[u]][0]==a[bob[v]][0]) addedge(u,v+n),addedge(v,u+n);
                if(a[bob[u]][0]==a[bob[v]][1]) addedge(u,v),addedge(v+n,u+n);
                if(a[bob[u]][1]==a[bob[v]][0]) addedge(u+n,v+n),addedge(v,u);
                if(a[bob[u]][1]==a[bob[v]][1]) addedge(u+n,v),addedge(v+n,u);
            }
            else
            {
                if(bob[u]==bob[v])
                {
                    addedge(u,v);
                    addedge(v,u);
                    addedge(u+n,v+n);
                    addedge(v+n,u+n);
                }
                else
                {   //注意这里 只能唯一的选择一条边等价于不能选其他边
                    if(a[bob[u]][0]==a[bob[v]][0]) addedge(u+n,u),addedge(v+n,v);
                    if(a[bob[u]][0]==a[bob[v]][1]) addedge(u+n,u),addedge(v,v+n);
                    if(a[bob[u]][1]==a[bob[v]][0]) addedge(u,u+n),addedge(v+n,v);
                    if(a[bob[u]][1]==a[bob[v]][1]) addedge(u,u+n),addedge(v,v+n);
                }
            }
        }
        Twosat();
        printf("Case #%d: ",++test);
        if(flag) printf("yes\n");
        else printf("no\n");
    }
    return 0;
}