3551: [ONTAK2010]Peaks加强版
3551: [ONTAK2010]Peaks加强版
https://www.lydsy.com/JudgeOnline/problem.php?id=3551
分析：
kruskal重构树 + 倍增 + 主席树。
首先建立kruskal重构树，那么查询就变成了，在kruskal重构树上找倍增找到最上面的权值小于x的点（节点的权值为原图的边权），那么这棵树内的所有点都可以在经过权值小于x的点相互到达，所以在这棵树内查询第k大即可。dfs序后，变成序列上的问题，查询区间的第k大。
代码：
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<cctype> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<map> 11 #define fi(s) freopen(s,"r",stdin); 12 #define fo(s) freopen(s,"w",stdout); 13 using namespace std; 14 typedef long long LL; 15 16 inline int read() { 17 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 18 for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 19 } 20 21 const int N = 100005; 22 23 struct Edge{ 24 int u, v, w; 25 bool operator < (const Edge &A) const { 26 return w < A.w; 27 } 28 }e[500005]; 29 int fa[N << 1], f[N << 1][19], mx[N << 1], st[N << 1], pos[N << 1], en[N << 1], val[N], disc[N]; // mx[N<<1] 30 int sum[N * 18], ls[N * 18], rs[N * 18], Root[N << 1]; // 乘18，不要17 31 int n, m, Q, tot, Time_Index, tot_node; 32 vector<int> T[N << 1]; 33 34 #define lson l, mid, rt << 1 35 #define rson mid + 1, r, rt << 1 | 1 36 37 int find(int x) { 38 return x == fa[x] ? x : fa[x] = find(fa[x]); 39 } 40 void Kruskal() { 41 for (int i=1; i<=n+n; ++i) fa[i] = i; 42 sort(e + 1, e + m + 1); 43 int cntedge = 0; tot = n; 44 for (int i=1; i<=m; ++i) { 45 int u = find(e[i].u), v = find(e[i].v); 46 if (u == v) continue; 47 ++tot; 48 fa[u] = tot, fa[v] = tot; 49 f[u][0] = tot, f[v][0] = tot; 50 mx[tot] = e[i].w; 51 T[tot].push_back(u), T[tot].push_back(v); 52 // if (++cntedge == n + n - 1) break; // n + n - 1 53 } 54 } 55 void dfs(int u,int fa) { 56 st[u] = ++Time_Index; 57 pos[Time_Index] = u; 58 for (int sz=T[u].size(),i=0; i<sz; ++i) { 59 int v = T[u][i]; 60 if (v == fa) continue; 61 dfs(v, u); 62 } 63 en[u] = Time_Index; 64 } 65 void update(int l,int r,int &rt,int last,int p) { 66 rt = ++tot_node; 67 sum[rt] = sum[last] + 1; 68 ls[rt] = ls[last], rs[rt] = rs[last]; 69 if (l == r) return ; 70 int mid = (l + r) >> 1; 71 if (p <= mid) update(l, mid, ls[rt], ls[last], p); 72 else update(mid + 1, r, rs[rt], rs[last], p); 73 } 74 int query(int l,int r,int Head,int Tail,int k) { 75 if (sum[Tail] - sum[Head] < k) return -1; 76 if (l == r) return l; 77 int mid = (l + r) >> 1, cnt = sum[ls[Tail]] - sum[ls[Head]]; 78 if (cnt >= k) return query(l, mid, ls[Head], ls[Tail], k); 79 else return query(mid + 1, r, rs[Head], rs[Tail], k - cnt); 80 } 81 int main() { 82 n = read(), m = read(), Q = read(); 83 for (int i=1; i<=n; ++i) val[i] = read(), disc[i] = val[i]; 84 for (int i=1; i<=m; ++i) 85 e[i].u = read(), e[i].v = read(), e[i].w = read(); 86 87 sort(disc + 1, disc + n + 1); 88 int cnt = 1; 89 for (int i=2; i<=n; ++i) if (disc[i] != disc[cnt]) disc[++cnt] = disc[i]; // i=2 90 for (int i=1; i<=n; ++i) val[i] = lower_bound(disc + 1, disc + cnt + 1, val[i]) - disc; 91 92 Kruskal(); 93 dfs(tot, 0); 94 95 for (int j=1; j<=18; ++j) 96 for (int i=1; i<=tot; ++i) f[i][j] = f[f[i][j-1]][j-1]; 97 98 for (int i=1; i<=tot; ++i) { 99 if (pos[i] > n) Root[i] = Root[i - 1]; 100 else update(1, cnt, Root[i], Root[i - 1], val[pos[i]]); // val[pos[i]] !!! 101 } 102 103 int lastans = 0; 104 while (Q--) { 105 int v = read(), x = read(), k = read(); 106 if (lastans != -1) v ^= lastans, x ^= lastans, k ^= lastans; 107 for (int i=18; i>=0; --i) 108 if (f[v][i] && mx[f[v][i]] <= x) v = f[v][i]; 109 int l = st[v], r = en[v]; 110 k = sum[Root[r]] - sum[Root[l]] - k + 1; 111 if (k <= 0) { puts("-1"); lastans = -1; continue; } 112 int p = query(1, cnt, Root[l], Root[r], k); 113 if (p == -1) { puts("-1"); lastans = -1; continue; } 114 printf("%d\n",lastans = disc[p]); 115 } 116 return 0; 117 }
