強制在線。

在kruskal重構樹上線段樹合並即可。

題意有毒，ans = -1的時候下一次不異或。

  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 }

bzoj3551 Peaks加強版