bzoj3545 Peaks

题意：多次求从点x出发经过边权不超过k的边能走到的点中第k大的权值。

解：离线排序 + 并查集 + 线段树合并。

题面有锅...是第k大的权值不是第k大的山。

 #include <cstdio>
 #include <algorithm>

 const int N = , M = , V = ;

 struct Edge {
     int x, y, h;
     inline bool operator <(const Edge &w) const {
         return h < w.h;
     }
 }edge[M];

 struct Ask {
     int x, h, k, id;
     inline bool operator <(const Ask &w) const {
         return h < w.h;
     }
 }ask[M];

 int fa[N], rt[N], ls[V], rs[V], sum[V], siz[N], tot, ans[M], X[N], val[N];

 int find(int x) {
     if(x == fa[x]) {
         return x;
     }
     return fa[x] = find(fa[x]);
 }

 int merge(int x, int y) {
     if(!x || !y) {
         return x | y;
     }
     int o = ++tot;
     sum[o] = sum[x] + sum[y];
     ls[o] = merge(ls[x], ls[y]);
     rs[o] = merge(rs[x], rs[y]);
     return o;
 }

 int query(int k, int l, int r, int o) {
     if(l == r) {
         return r;
     }
     int mid = (l + r) >> ;
     if(k > sum[rs[o]]) {
         return query(k - sum[rs[o]], l, mid, ls[o]);
     }
     else {
         return query(k, mid + , r, rs[o]);
     }
 }

 inline void Xmerge(int x, int y) {
     x = find(x);
     y = find(y);
     if(x == y) {
         return;
     }
     fa[y] = x;
     siz[x] += siz[y];
 //    printf("siz %d += %d = %d \n", x, siz[y], siz[x]);
     rt[x] = merge(rt[x], rt[y]);
     return;
 }

 void insert(int p, int l, int r, int &o) {
     if(!o) {
         o = ++tot;
     }
     sum[o] = ;
     if(l == r) {
         return;
     }
     int mid = (l + r) >> ;
     if(p <= mid) {
         insert(p, l, mid, ls[o]);
     }
     else {
         insert(p, mid + , r, rs[o]);
     }
     return;
 }

 int main() {
     int n, m, q;
     scanf("%d%d%d", &n, &m, &q);
     for(int i = ; i <= n; i++) {
         scanf("%d", &val[i]);
         X[i] = val[i];
     }
     std::sort(X + , X + n + );
     int xx = std::unique(X + , X + n + ) - X - ;
     for(int i = ; i <= n; i++) {
         val[i] = std::lower_bound(X + , X + xx + , val[i]) - X;
         insert(val[i], , xx, rt[i]);
         fa[i] = i; siz[i] = ;
     }
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].h);
     }
     for(int i = ; i <= q; i++) {
         scanf("%d%d%d", &ask[i].x, &ask[i].h, &ask[i].k);
         ask[i].id = i;
     }
     std::sort(edge + , edge + m + );
     std::sort(ask + , ask + q + );

     int p = ;
     for(int i = ; i <= q; i++) {
         while(p <= m && edge[p].h <= ask[i].h) {
             Xmerge(edge[p].x, edge[p].y);
             p++;
 //            printf("Xmerge %d %d \n", edge[p].x, edge[p].y);
         }
         int x = find(ask[i].x);
 //        printf("x = %d \n", x);
 //        printf("%d < %d \n", siz[x], ask[i].k);
         if(siz[x] < ask[i].k) ans[ask[i].id] = -;
         else ans[ask[i].id] = X[query(ask[i].k, , xx, rt[x])];
     }
     for(int i = ; i <= q; i++) {
         printf("%d\n", ans[i]);
     }
     return ;
 }

AC代码