二次联通门 : luoguP1046
/*这个题好难.....由苹果树可知这应该是个树结构的题所以很自然的想到了用树链剖分来搞一下连边 最后查询以1为根节点的子树的权值和...从前闲的没事写着玩... */ #include <cstdio> #define Max 3300void read (int &now) {now = 0;char word = getchar ();while (word > '9' || word < '0')word = getchar ();while (word >= '0' && word <= '9'){now = now * 10 + word - '0';word = getchar ();} }static int Hight;struct Edge {int to;int next; };int Edge_Count; int edge_list[Max]; Edge edge[Max];inline void AddEdge (int from, int to) {Edge_Count++;edge[Edge_Count].to = to;edge[Edge_Count].next = edge_list[from];edge_list[from] = Edge_Count;Edge_Count++;edge[Edge_Count].to = from;edge[Edge_Count].next = edge_list[to];edge_list[to] = Edge_Count; }struct Point {int size;int deep;int dis;int father;int tree_number;int End; };struct Tree {int l;int r;int dis;int Mid; };Tree tree[Max]; Point point[Max]; int Count;void Dfs_1 (int now, int father) {int pos = Count++;point[now].father = father;point[now].deep = point[father].deep + 1;for (int i = edge_list[now]; i; i = edge[i].next){if (edge[i].to == father)continue;Dfs_1 (edge[i].to, now);}point[now].size = Count - pos; }int tree_dis[Max];void Dfs_2 (int now, int chain) {int pos = 0;point[now].tree_number = ++Count;tree_dis[Count] = point[now].dis;for (int i = edge_list[now]; i; i = edge[i].next){if (point[edge[i].to].tree_number)continue;if (point[edge[i].to].size > point[pos].size)pos = edge[i].to;}if (pos)Dfs_2 (pos, chain);for (int i = edge_list[now]; i; i = edge[i].next){if (point[edge[i].to].tree_number)continue;Dfs_2 (edge[i].to, edge[i].to);}point[now].End = Count; }void Tree_Build (int l, int r, int now) {tree[now].l = l;tree[now].r = r;if (l == r){tree[now].dis = tree_dis[++Count] <= (30 + Hight) ? 1 : 0;return ;}tree[now].Mid = (l + r) >> 1;Tree_Build (l, tree[now].Mid, now << 1);Tree_Build (tree[now].Mid + 1, r, now << 1 | 1);tree[now].dis = tree[now << 1].dis + tree[now << 1 | 1].dis; }int Tree_Query (int l, int r, int now) {if (tree[now].l == l && tree[now].r == r)return tree[now].dis;if (r <= tree[now].Mid)return Tree_Query (l, r, now << 1);else if (l > tree[now].Mid)return Tree_Query (l, r, now << 1 | 1);elsereturn Tree_Query (l, tree[now].Mid, now << 1) + Tree_Query (tree[now].Mid + 1, r, now << 1 | 1); }int main (int argc, char *argv[]) {for (int i = 1; i <= 10; i++)read (point[i].dis);read (Hight);for (int i = 1; i < 10; i++)AddEdge (i, i + 1);Count = 0;Dfs_1 (1, 1);Count = 0;Dfs_2 (1, 0);Count = 0;Tree_Build (1, 10, 1);printf ("%d", Tree_Query (point[1].tree_number, point[1].End, 1));return 0; }
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