题目

CF576E

分析：

从前天早上肝到明天早上qwq其实颓了一上午MC ，自己瞎yy然后1A，写篇博客庆祝一下。

首先做这题之前推荐一道很相似的题：【BZOJ4025】二分图（可撤销并查集+线段树分治）

大力每个颜色维护一个并查集，就很像上面那道题了。但是存在一个问题：在处理线段树区间\([l,r]\)时，可能并不知道\(l\)处的修改是否成功，所以不知道\(l\)处修改的边具体是什么颜色的。

我的解决方案是：处理区间\([l,r]\)时忽略\(l\)处修改的边。先向左子树递归，递归到叶子时判断本次修改颜色能否成功。然后回溯后向右子树递归前将这条边加入。

代码：

solve函数的第四个参数表示现在是否已经忽略了\(l\)处修改的边。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <queue>
#include <bitset>
#include <stack>using namespace std;namespace zyt
{template<typename T>inline void read(T &x){char c;bool f = false;x = 0;doc = getchar();while (c != '-' && !isdigit(c));if (c == '-')f = true, c = getchar();dox = x * 10 + c - '0', c = getchar();while (isdigit(c));if (f)x = -x;}template<typename T>inline void write(T x){static char buf[20];char *pos = buf;if (x < 0)putchar('-'), x = -x;do*pos++ = x % 10 + '0';while (x /= 10);while (pos > buf)putchar(*--pos);}inline void write(const char *const s){printf("%s", s);}const int N = 5e5 + 10, B = 19, K = 51;int n, m, k, q, head[1 << (B + 1) | 11], ecnt;struct UFS{int fa[N], rk[N];bitset<N> dis;struct node{UFS &ufs;int x, y, fa, rk, dis;};static stack<node> sta;inline void init(){for (int i = 0; i < N; i++)fa[i] = i, rk[i] = 1;dis = 0U;}int f(const int x){return x == fa[x] ? x : f(fa[x]);}int dist(const int x){return x == fa[x] ? dis[x] : dist(fa[x]) ^ dis[x];}inline bool merge(const int u, const int v){int x = f(u), y = f(v);if (x == y)return dist(u) ^ dist(v);if (rk[x] > rk[y])swap(x, y);sta.push((node){*this, x, y, fa[x], rk[y], dis[x]});fa[x] = y, dis[x] = dis[x] ^ dist(u) ^ dist(v) ^ 1;if (rk[x] == rk[y])++rk[y];return true;}static inline int set_undo(){return sta.size();}static inline void undo(const int bck){while (sta.size() > bck){UFS &now = sta.top().ufs;now.fa[sta.top().x] = sta.top().fa;now.rk[sta.top().y] = sta.top().rk;now.dis[sta.top().x] = sta.top().dis;sta.pop();}}}ufs[K];stack<UFS::node> UFS::sta;struct edge{int id, next;}e[N * B];inline void add(const int a, const int b){e[ecnt] = (edge){b, head[a]}, head[a] = ecnt++;}struct ed{int u, v;}arr[N];struct node{int ed, col;}mdf[N];int pre[N], nxt[N], last[N];namespace Segment_Tree{void insert(const int rot, const int lt, const int rt, const int ls, const int rs, const int id){if (ls <= lt && rt <= rs){add(rot, id);return;}int mid = (lt + rt) >> 1;if (ls <= mid)insert(rot << 1, lt, mid, ls, rs, id);if (rs > mid)insert(rot << 1 | 1, mid + 1, rt, ls, rs, id);}void solve(const int rot, const int lt, const int rt, bool flag){int mid = (lt + rt) >> 1;int bck = UFS::set_undo();bool f = false;for (int i = head[rot]; ~i; i = e[i].next){int now = e[i].id;if (lt == now){flag = true;continue;}UFS &u = ufs[mdf[now].col];ed &edg = arr[mdf[now].ed];if (mdf[now].col)u.merge(edg.u, edg.v);}if (lt == rt){if (!ufs[mdf[lt].col].merge(arr[mdf[lt].ed].u, arr[mdf[lt].ed].v))mdf[lt].col = mdf[pre[lt]].col, write("NO");elsewrite("YES");putchar('\n');}else{solve(rot << 1, lt, mid, f | flag);if (f | flag)ufs[mdf[lt].col].merge(arr[mdf[lt].ed].u, arr[mdf[lt].ed].v);solve(rot << 1 | 1, mid + 1, rt, false);}UFS::undo(bck);}}int work(){using namespace Segment_Tree;memset(head, -1, sizeof(head));read(n), read(m), read(k), read(q);for (int i = 1; i <= k; i++)ufs[i].init();for (int i = 1; i <= m; i++)read(arr[i].u), read(arr[i].v);for (int i = 1; i <= q; i++){read(mdf[i].ed), read(mdf[i].col);nxt[i] = q + 1;}for (int i = 1; i <= q; i++){if (last[mdf[i].ed])pre[i] = last[mdf[i].ed];last[mdf[i].ed] = i;}for (int i = q; i > 0; i--)nxt[pre[i]] = i;for (int i = 1; i <= q; i++)insert(1, 1, q, i, nxt[i] - 1, i);solve(1, 1, q, false);return 0;}
}
int main()
{return zyt::work();
}