The 2010 ACM-ICPC Asia Chengdu Regional Contest - G Go Deeper 二分+2-SAT

The 2010 ACM-ICPC Asia Chengdu Regional Contest - G Go Deeper 二分+2-SAT

Go Deeper Time Limit: 2 Seconds      Memory Limit: 65536 KB

Here is a procedure's pseudocode:

	   go(int dep, int n, int m)
begin
output the value of dep.
if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
end

 

In this code n is an integer. a, b, c and x are 4 arrays of integers. The index of array always starts from 0. Array a and b consist of non-negative integers smaller than n. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, b and c are m while the length of array x is n.

Given the elements of array a, b, and c, when we call the procedure go(0, n , m) what is the maximal possible value does the procedure output?

Input

There are multiple test cases. The first line of input is an integer T (0 < T ≤ 100), indicating the number of test cases. Then T test cases follow. Each case starts with a line of 2 integers n and m (0 < n ≤ 200, 0 < m ≤ 10000). Then m lines of 3 integers follow. The i-th(1 ≤ i ≤ m) line of them are a_i-1 ,b_i-1 and c_i-1 (0 ≤ a_i-1, b_i-1 < n, 0 ≤ c_i-1 ≤ 2).

Output

For each test case, output the result in a single line.

Sample Input

3
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2

 

Sample Output

1
1
2

解法：

看到x为二进制数组就很敏感了，然后又看到c的值只能取0,1,2。很快想到了2-sat

t1=a[i],t2=b[i]

如果x[t1]+x[t2]!=0，那么就是说t1的0和t2的0冲突，添加t10->t21和t20->t11 

如果x[t1]+x[t2]!=2，那么就是说t1的1和t2的1冲突，添加t11->t20和t21->t10 

如果x[t1]+x[t2]!=1，那么就是说t1的0和t2的1冲突，以及t1的1和t2的0冲突，添加t10->t20和t21->t11 以及t11->t21和t20->t10 
贴代码

   
   
   
   

    
    
    
     1
    
    
    
    
    
    
    
    # include 
    
    
    
    <
    
    
    
    cstdio
    
    
    
    >
    
    
    
    

    
    
    
     2
    
    
    
    # include 
    
    
    
    <
    
    
    
    vector
    
    
    
    >
    
    
    
    

    
    
    
     3
    
    
    
    # include 
    
    
    
    <
    
    
    
    cstring
    
    
    
    >
    
    
    
    

    
    
    
     4
    
    
    
    
    
    
    
    using
    
    
    
     
    
    
    
    namespace
    
    
    
     std;

    
    
    
     5
    
    
    
    
    
    
    
    int
    
    
    
     n,m;

    
    
    
     6
    
    
    
    vector
    
    
    
    <
    
    
    
    int
    
    
    
    >
    
    
    
     g[
    
    
    
    500
    
    
    
    ];

    
    
    
     7
    
    
    
    
    
    
    
    int
    
    
    
     a[
    
    
    
    10005
    
    
    
    ],b[
    
    
    
    10005
    
    
    
    ],c[
    
    
    
    10005
    
    
    
    ];

    
    
    
     8
    
    
    
    
    
    
    
    int
    
    
    
     low[
    
    
    
    500
    
    
    
    ],dfn,color[
    
    
    
    500
    
    
    
    ],stack[
    
    
    
    500
    
    
    
    ],top,co;

    
    
    
     9
    
    
    
    
    
    
    
    void
    
    
    
     tarjan(
    
    
    
    int
    
    
    
     pos)

    
    
    
    10
    
    
    
    
    
    
    
    
    
    
    
    {
11   int now=dfn++;
12   low[pos]=now;
13   stack[++top]=pos;
14   for(int i=0;i<g[pos].size();i++)
15   {
16      if(low[g[pos][i]]==-1) tarjan(g[pos][i]);
17      if(low[g[pos][i]]<low[pos]) low[pos]=low[g[pos][i]];
18   }
19   if(low[pos]>=now)
20   {
21       do
22       {
23          color[stack[top]]=co;
24          low[stack[top]]=2*n;
25       }while(stack[top--]!=pos);
26       co++;
27   }
28   
29}
    
    
    
    

    
    
    
    30
    
    
    
    
    
    
    
    bool
    
    
    
     chk(
    
    
    
    int
    
    
    
     num)

    
    
    
    31
    
    
    
    
    
    
    
    
    
    
    
    {
32   for(int i=0;i<(n<<1);i++)
33      g[i].clear();
34   for(int i=0;i<=num;i++)
35      switch(c[i])
36      {
37         case 0:
38              g[a[i]*2].push_back(b[i]*2+1);
39              g[b[i]*2].push_back(a[i]*2+1);
40              break;
41         case 1:
42              g[a[i]*2].push_back(b[i]*2);
43              g[b[i]*2+1].push_back(a[i]*2+1);
44              g[a[i]*2+1].push_back(b[i]*2+1);
45              g[b[i]*2].push_back(a[i]*2);
46              break;
47         case 2:
48              g[a[i]*2+1].push_back(b[i]*2);
49              g[b[i]*2+1].push_back(a[i]*2);
50              break;
51      };
52   memset(low,-1,sizeof(low));
53   dfn=co=0;
54   memset(color,-1,sizeof(color));
55   for(int i=0;i<2*n;i++)
56     if(low[i]==-1)
57     {
58       top=-1;
59       tarjan(i);
60     }
61   for(int i=0;i<n;i++)
62     if(color[2*i]==color[2*i+1])
63       return false;
64   return true;
65}
    
    
    
    

    
    
    
    66
    
    
    
    
    
    
    
    int
    
    
    
     main()

    
    
    
    67
    
    
    
    
    
    
    
    
    
    
    
    {
68    int test;
69    scanf("%d",&test);
70    while(test--)
71    {
72        scanf("%d%d",&n,&m);
73        for(int i=0;i<m;i++)
74          scanf("%d%d%d",a+i,b+i,c+i);
75        int s=0,e=m-1;
76        while(s<=e)
77        {
78           int mid=(s+e)>>1;
79           if(chk(mid)) s=mid+1;
80           else e=mid-1;
81        }
82        printf("%d\n",s);
83    }
84    return 0;
85}
    
    
    
    

    
    
    
    86
    
    
    
    

    
    
    
    87