题干：
链接：https://ac.nowcoder.com/acm/contest/883/A
来源：牛客网
You are given an undirected graph with N\ N N vertices and M\ M M edges. The edges are numbered from 1\ 1 1 to M\ M M . Denote the set S(x)\ S(x) S(x) as: All the vertices we can reach from vertex x\ x x by exactly one edge.
You are supposed to deal with Q\ Q Q operations of the following two types:
- (1,l,r)\ (1,l,r) (1,l,r)-- reverse the status of edges numbered between l\ l l to r\ r r (1≤l≤r≤M)(1 \leq l \leq r \leq M)(1≤l≤r≤M) . i.e. Delete the edge if it is in the graph, otherwise add it to the graph.
- (2,u,v)\ (2,u,v) (2,u,v) -- ask whether S(u)\ S(u) S(u) and S(v)\ S(v) S(v) are exactly the same (1≤u≤v≤N)(1 \leq u \leq v \leq N)(1≤u≤v≤N).
Note that all the M\ M M edges are in the graph at the beginning.
输入描述:
The input contains multiple cases. The first line of the input contains a single positive integer T\ T T , the number of cases.For each case, the first line contains two integers N (1≤N≤100000)N\ (1 \leq N \leq 100000)N (1≤N≤100000) and M (1≤M≤200000)M\ (1 \leq M \leq 200000)M (1≤M≤200000), the number of vertices and edges in the graph. In the following M\ M M lines, the i\ i i-th line contains two integers ui,vi (1≤ui,vi≤N)u_i,v_i \ (1 \le u_i,v_i \le N)ui,vi (1≤ui,vi≤N) , describing the the i\ i i-th edge (ui,vi)(u_i,v_i)(ui,vi) . Each edge appears in the input at most once. The (M+2)\ (M+2) (M+2)-th line contains a integer Q (1≤Q≤200000)Q\ (1 \leq Q \leq 200000)Q (1≤Q≤200000) , the number of operations. In the following Q\ Q Q lines, each line contains three integers, describing an operation.The total sum of N\ N N over all cases does not exceed 150000\ 150000 150000. The total sum of M\ M Mover all cases does not exceed 800000\ 800000 800000. The total sum of Q\ Q Q over all cases does not exceed 600000\ 600000 600000.
输出描述:
For each case, print a string in a line. The length of the string should equal the number of operations of type 2\ 2 2. If the answer is yes, the i\ i i-th character of the string should be `1', otherwise it should be `0'. Check the samples for more details.
示例1
输入
复制
1 5 4 1 2 1 3 4 2 4 3 3 2 1 4 1 1 1 2 1 2
输出
复制
10
题目大意：
一个n个点m条边的的图，2种操作：
1 l r 表示从读入的编号为l的边到第r条边，如果有这条边就删掉，否则就加上
2 x y 表示问与点x直接相连的点集是否与y直接相连的点集相同
解题报告：
Hash数组是每个点映射成一个值。
L 和 R 数组对应每个块的左右编号。
O代表单点更新的点的Hash值，laz代表整块的翻转次数的奇偶数。
B数组是预处理数组，在操作的时候不更新，预处理每个块中的每个点向外连边的Hash值。
将输入的边编号然后对边分块，如果要反转的两个区间不在相邻块内或者相同块，那么暴力l块和暴力r块，块间使用laz标记是否翻转。很优秀的想法。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1e5 + 5;
int blk[MAX];
int a[MAX<<1],b[MAX<<1];
int n,m,q,tot,Block;
int L[505],R[505],B[505][MAX],laz[505],O[MAX],Hash[MAX];
int main()
{for(int i = 1; i<=1e5; i++) Hash[i] = rand(); int t;cin>>t;while(t--) {tot=0;scanf("%d%d",&n,&m);for(int i = 1; i<=n; i++) O[i] = 0;for(int i = 1; i<=m; i++) cin>>a[i]>>b[i];Block = sqrt(m);for(int i = 1; i<=m; i+=Block) {L[++tot] = i;R[tot] = min(m,i+Block-1);for(int j = 1; j<=n; j++) B[tot][j] = 0;for(int j = L[tot]; j<=R[tot]; j++) B[tot][a[j]] ^= Hash[b[j]],B[tot][b[j]] ^= Hash[a[j]];laz[tot] = 0; }scanf("%d",&q);while(q--) {int op,l,r; scanf("%d%d%d",&op,&l,&r);if(op == 2) {int x = O[l],y = O[r];for(int i = 1; i<=tot; i++) {if(!laz[i]) x ^= B[i][l],y ^= B[i][r];}printf("%d",x == y);}else {int B1 = (l+Block-1)/Block,B2 = (r+Block-1)/Block;if(B2-B1>=2) {for(int i = B1+1; i<=B2-1; i++) laz[i] ^= 1;for(int i = l; i<=R[B1]; i++) O[a[i]] ^= Hash[b[i]],O[b[i]] ^= Hash[a[i]];for(int i = L[B2]; i<=r; i++) O[a[i]] ^= Hash[b[i]],O[b[i]] ^= Hash[a[i]];}else {for(int i = l; i<=r; i++) O[a[i]] ^= Hash[b[i]],O[b[i]] ^= Hash[a[i]];} }}puts("");}return 0 ;
}
