Description

To an undirected graph, while white and black colors, now you have a bunch of anti-color brush, brush can start from any point, after several sides back to the beginning.
Now to ask how many anti-color brush can at least make this picture all the edges are white.
For some reason, the edge color will change, so. .
We need to support the following:
. 1 x x sides of the article in reverse color (m-numbered. 1 ~ 0)
2 querying the current drawing color requires a minimum number of counter-brush

Input

The first two rows represent integers nm This map contains n points m edges
next m lines each uvc represents an integer of 3 undirected edges and the color of this edge (0 to 1 white black)
Next, a q q represents an integer of operations has
the following operation described above behavior q

Output

For each line a query output integer
represents the minimum required number of brush reverse color scheme output if no valid -1

Sample Input

6 6
1 2 1
2 3 1
1 3 1
4 5 1
5 6 1
4 6 1
14
2
1 0
2
1 1
1 2
2
1 3
1 4
1 5
2
1 3
1 4
1 5
2

Sample Output

2
-1
1
0
1

Hint

100% n, m, q <= 1000000, c <2, no multiple edges loopback

answer

First Title solvable if and only if the number of black bars are connected to each point of an even number
The number of black bars if the communication blocks is equal to text answer is present
Proof: For a communication block, assumed that the current paths respectively selected from two start u and v disjoint
We can u go through as a starting point of the path, and then walked v from u, v path from the beginning of the walk
And finally back to u to merge these two paths

Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int N=1000005;
 6 int n,m,cnt,now,q,f[N],d[N],col[N];
 7 struct edge{int x,y,c;}e[N];
 8 int find(int x) { return f[x]==x?x:f[x]=find(f[x]); }
 9 int main()
10 {
11     scanf("%d%d",&n,&m);
12     for (int i=1;i<=n;i++) f[i]=i;
13     for (int i=1,x,y,z;i<=m;i++)
14     {
15         scanf("%d%d%d",&x,&y,&z),e[i].x=x,e[i].y=y,e[i].c=z;
16         if (z) now-=d[x]+d[y],d[x]^=1,d[y]^=1,now+=d[x]+d[y];
17         if (find(x)!=find(y)) f[find(x)]=find(y);
18     }
19     for (int i=1;i<=m;i++)
20         if (e[i].c)
21         {
22             int x=find(e[i].x);
23             if (!col[x]) cnt++;
24             col[x]++;
25         }
26     scanf("%d",&q);
27     for (int op,x;q;q--)
28     {
29         scanf("%d",&op);
30         if (op==2) printf("%d\n",now?-1:cnt);
31         else
32         {
33             scanf("%d",&x),x++;
34             now-=d[e[x].x]+d[e[x].y],d[e[x].x]^=1;d[e[x].y]^=1,now+=d[e[x].x]+d[e[x].y];
35             if (e[x].c)
36             {
37                 e[x].c=0,col[find(e[x].x)]--;
38                 if (!col[find(e[x].x)]) cnt--;
39             }
40             else
41             {
42                 e[x].c=1;
43                 if (!col[find(e[x].x)]) cnt++;
44                 col[find(e[x].x)]++;
45             }
46         }
47     }
48     return 0;
49 }

[Euler] [disjoint-set] Bzoj P3706 reverse color brush