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 ", 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 ", lastans); 143 } 144 } 145 146 return 0; 147 }