bzoj3551 Peaks加强版
作者：互联网
强制在线。
在kruskal重构树上线段树合并即可。
题意有毒，ans = -1的时候下一次不异或。
1 /**************************************************************
2 Problem: 3551
3 Language: C++
4 Result: Accepted
5 Time:19204 ms
6 Memory:126608 kb
7 ****************************************************************/
8
9 #include <cstdio>
10 #include <algorithm>
11 #include <cstring>
12
13 const int N = 100010, M = 500010, V = 8500010;
14
15 struct Edge {
16 int x, y, h;
17 inline bool operator <(const Edge &w) const {
18 return h < w.h;
19 }
20 }edge[M];
21
22 int val[N], X[N];
23 int ls[V], rs[V], sum[V], tot;
24 int num, fa[N * 2][20], h[N * 2], rt[N * 2], siz[N * 2], pw[N * 2];
25
26 namespace ufs {
27 int fa[N * 2];
28 inline void init(int n) {
29 for(int i = 1; i <= n; i++) {
30 fa[i] = i;
31 }
32 return;
33 }
34 int find(int x) {
35 if(x == fa[x]) return x;
36 return fa[x] = find(fa[x]);
37 }
38 inline void merge(int x, int y) {
39 fa[find(y)] = find(x);
40 return;
41 }
42 inline bool check(int x, int y) {
43 return find(x) == find(y);
44 }
45 }
46
47 int merge(int x, int y) {
48 if(!x || !y) return x | y;
49 int o = ++tot;
50 sum[o] = sum[x] + sum[y];
51 ls[o] = merge(ls[x], ls[y]);
52 rs[o] = merge(rs[x], rs[y]);
53 return o;
54 }
55
56 void add(int p, int l, int r, int &o) {
57 if(!o) o = ++tot;
58 sum[o] = 1;
59 if(l == r) return;
60 int mid = (l + r) >> 1;
61 if(p <= mid) add(p, l, mid, ls[o]);
62 else add(p, mid + 1, r, rs[o]);
63 return;
64 }
65
66 int ask(int k, int l, int r, int o) {
67 //printf("%d %d %d %d \n", k, l, r, o);
68 if(l == r) return r;
69 int mid = (l + r) >> 1;
70 if(k <= sum[rs[o]]) return ask(k, mid + 1, r, rs[o]);
71 else return ask(k - sum[rs[o]], l, mid, ls[o]);
72 }
73
74 inline int getPos(int x, int H) {
75 int t = pw[num];
76 while(t >= 0 && fa[x][0] && h[fa[x][0]] <= H) {
77 if(fa[x][t] && h[fa[x][t]] <= H) {
78 x = fa[x][t];
79 }
80 t--;
81 }
82 return x;
83 }
84
85 int main() {
86
87 //freopen("in.in", "r", stdin);
88 //freopen("my.out", "w", stdout);
89
90 int n, m, q;
91 scanf("%d%d%d", &n, &m, &q);
92 num = n;
93 ufs::init(n + n);
94 for(int i = 1; i <= n; i++) {
95 scanf("%d", &val[i]);
96 X[i] = val[i];
97 }
98 std::sort(X + 1, X + n + 1);
99 int xx = std::unique(X + 1, X + n + 1) - X - 1;
100 for(int i = 1; i <= n; i++) {
101 val[i] = std::lower_bound(X + 1, X + xx + 1, val[i]) - X;
102 add(val[i], 1, xx, rt[i]);
103 siz[i] = 1;
104 }
105 for(int i = 1; i <= m; i++) {
106 scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].h);
107 }
108 std::sort(edge + 1, edge + m + 1);
109 // build tree
110 for(int i = 1; i <= m; i++) {
111 int x = edge[i].x, y = edge[i].y;
112 if(ufs::check(x, y)) {
113 continue;
114 }
115 ++num;
116 h[num] = edge[i].h;
117 x = ufs::find(x);
118 y = ufs::find(y);
119 rt[num] = merge(rt[x], rt[y]);
120 ufs::merge(num, x);
121 ufs::merge(num, y);
122 fa[x][0] = fa[y][0] = num;
123 siz[num] = siz[x] + siz[y];
124 }
125 for(int i = 2; i <= num; i++) {
126 pw[i] = pw[i >> 1] + 1;
127 }
128 for(int j = 1; j <= pw[num]; j++) {
129 for(int i = 1; i <= num; i++) {
130 fa[i][j] = fa[fa[i][j - 1]][j - 1];
131 }
132 }
133
134 int lastans = 0;
135 for(int i = 1, x, H, k; i <= q; i++) {
136 scanf("%d%d%d", &x, &H, &k);
137 if(lastans != -1) {x ^= lastans; H ^= lastans; k ^= lastans; }
138 int y = getPos(x, H);
139 if(siz[y] < k) puts("-1"), lastans = -1;
140 else {
141 lastans = X[ask(k, 1, xx, rt[y])];
142 printf("%d\n", lastans);
143 }
144 }
145
146 return 0;
147 }
AC代码
标签：Peaks,int,19204,fa,题意,&&,include,bzoj3551,加强版 来源： https://www.cnblogs.com/huyufeifei/p/10415064.html