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 ai-1 ,bi-1 and ci-1 (0 ≤ ai-1, bi-1 < n, 0 ≤ ci-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為二進制數組就很敏感了，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，那么就是說t1的0和t2的0沖突，添加t10->t21和t20->t11
如果x[t1]+x[t2]!=2，那么就是說t1的1和t2的1沖突，添加t11->t20和t21->t10
如果x[t1]+x[t2]!=1，那么就是說t1的0和t2的1沖突，以及t1的1和t2的0沖突，添加t10->t20和t21->t11 以及t11->t21和t20->t10
貼代碼
1
# include <cstdio>
2# include <vector>
3# include <cstring>
4using namespace std;
5int n,m;
6vector<int> g[500];
7int a[10005],b[10005],c[10005];
8int low[500],dfn,color[500],stack[500],top,co;
9void 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
}
30bool 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}
66int 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