可撤销并查集

之前没有写过可撤销并查集，这里整理一下.  

普通并查集是不支持撤销/断边操作的.  

但是如果加边顺序是 $(1,2,3,4,5)$, 断边顺序是 $(5,4,3,2,1)$ 的话是可以维护的.  

我们只需要用一个启发式合并的并查集加上栈来存储合并信息即可.  

这样做可行，是因为始终是向上合并，并且删除时是从上向下删除的.   

具体代码如下： 

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namespace ufs {     stack<int>S;     int dis[N], fa[N], size[N];      void init() {         for(int i=1;i<N;++i) {             dis[i] = 0, fa[i] = i, size[i] = 1;         }     }     int find(int x) {         return fa[x] == x ? fa[x] : find(fa[x]);     }     int dist(int x) {         return fa[x] == x ? dis[x] : dis[x] ^ dist(fa[x]);     }     void merge(int x, int y) {         int dx = dist(x);         int dy = dist(y);         x = find(x), y = find(y);          if(size[x] > size[y]) {             swap(x, y);             swap(dx, dy);          }         // size[y] > size[x]         // 令 f[x] -> y         S.push(x);         size[y] += size[x];          fa[x] = y, dis[x] = dx ^ dy ^ 1;      }     void undo(int si) {         // 撤销.          while(S.size() > si) {             int p = S.top(); S.pop();              size[fa[p]] -= size[p];              fa[p] = p, dis[p] = 0;             }       } };

　　

在撤销的时候删除 $\mathrm{x}$ 对 $\mathrm{fa[x]}$ 的影响就好了，这一般是较好维护的.  

1.二分图 /【模板】线段树分治

来源：luogu5787 

在这道题中，如果说没有删边操作则可以直接利用加权并查集维护奇偶性.  

然后有删边时间的话就利用线段树分治存边，利用加权并查集在线段树上合并和撤销即可.  

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#include <cstdio> #include <vector>  #include <cstring> #include <stack> #include <algorithm> #define N  200009 #define ll long long #define pb push_back #define ls now << 1 #define rs now << 1 | 1 #define setIO(s) freopen(s".in","r",stdin) using namespace std; namespace ufs {     stack<int>S;     int dis[N], fa[N], size[N];      void init() {         for(int i=1;i<N;++i) {             dis[i] = 0, fa[i] = i, size[i] = 1;         }     }     int find(int x) {         return fa[x] == x ? fa[x] : find(fa[x]);     }     int dist(int x) {         return fa[x] == x ? dis[x] : dis[x] ^ dist(fa[x]);     }     void merge(int x, int y) {         int dx = dist(x);         int dy = dist(y);         x = find(x), y = find(y);          if(size[x] > size[y]) {             swap(x, y);             swap(dx, dy);          }         // size[y] > size[x]         // 令 f[x] -> y         S.push(x);         size[y] += size[x];          fa[x] = y, dis[x] = dx ^ dy ^ 1;      }     void undo(int si) {         // 撤销.          while(S.size() > si) {             int p = S.top(); S.pop();              size[fa[p]] -= size[p];              fa[p] = p, dis[p] = 0;             }       } }; int n, m, K;   struct oper {     int x, y;      oper(int x=0,int y=0):x(x),y(y){}  }; vector<oper>G[N << 2];  void update(int l, int r, int now, int L, int R, oper v) {     if(l >= L && r <= R) {         G[now].pb(v);         return ;     }     int mid = (l + r) >> 1;     if(L <= mid) update(l, mid, ls, L, R, v);     if(R  > mid) update(mid + 1, r, rs, L, R, v);  } void dfs(int l, int r, int now, int flag) {     int cur = ufs::S.size();     if(!flag) {         // 若还没构成二分图，则可以加边.         for(int i = 0 ; i < G[now].size() ; ++ i) {             int x = G[now][i].x;             int y = G[now][i].y;              int px = ufs::find(x);             int py = ufs::find(y);             if(px != py) {                 ufs::merge(x, y);             }             else {                 int dx = ufs::dist(x);                 int dy = ufs::dist(y);                 if(dx ^ dy) {                     continue;                 }                 else {                     flag = 1;                     break;                 }             }         }     }     if(l == r) {         printf("%s\n", flag ? "No" : "Yes");          return ;     }     int mid = (l + r) >> 1;     dfs(l, mid, ls, flag);     dfs(mid + 1, r, rs, flag);      ufs::undo(cur); } int main() {     // setIO("input");     scanf("%d%d%d",&n,&m,&K);      ufs::init();     for(int i = 1; i <= m ; ++ i) {         int x, y, l, r;         scanf("%d%d%d%d",&x, &y, &l, &r);            update(0, K - 1, 1, l, r - 1, oper(x, y));      }     // 处理完了, 可以遍历了.      dfs(0, K - 1, 1, 0);     return 0; }