bzoj1935.园丁的烦恼 (离散化 && 树状数组)

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(SHOI2007) 平面内有N 个点，有 M 个查询，每次询问一个以 ( Ai, Bi ) 为左下角，( Ci, Di ) 为右上角的矩形内有多少个点

这种问题首先想到的应该是二维线段树或者树状数组，但是这道题中N， M 都是 500000，二维的肯定为超出空间限制，这能用一维。同时这道题中没有中途修改操作，所以可以离线做。把这些点离散化，询问的点也算在内，按 X 为第一关键字，是否为插入的点为第二关键字排序，这样就能做到先插入点，再查询。询问计算个数时，都按以( 0， 0 ) 为左下角的矩阵，用树状数组查询，这样能做到 O( logn )，之后和差一下就行。

#include <cstdio>#include <algorithm>using namespace std;#define lowbit(x) (x & (-x))const int MAX_N = 500005;struct node{int x, y, id, f;friend bool operator<(const node& a, const node& b){return a.x < b.x || (a.x == b.x && a.f < b.f); } // sort x}Q[MAX_N * 5];inline int read(){int ret = 0, f = 1; char c = getchar();while (!(c >= '0' && c <= '9')){ if (c == '-') f = -1; c = getchar(); }while (c >= '0' && c <= '9') ret = ret*10 + c-'0', c = getchar();return ret; }int N, M, x[MAX_N], y[MAX_N], st[MAX_N * 3];int a[MAX_N], b[MAX_N], c[MAX_N], d[MAX_N];int tot = 0, all = 0, tree[MAX_N][5]; // lisan, sum, ansint s[MAX_N * 3]; // treeint got(int x){int l = 1, r = tot;while(l <= r){int mid = (l+r) >> 1;if(st[mid] == x) return mid;  else if(st[mid] > x) r = mid - 1;  else l = mid + 1;}}void tree_add(int x){for (int i=x; i<=tot; i += lowbit(i))s[i] += 1;}int tree_query(int x){int ret = 0;for(int i = x; i; i -= lowbit(i))ret += s[i];return ret;}void work(void){sort(Q+1, Q+all+1);for (int i=1; i<=all; i++){printf("%d %d %d %d\n", Q[i].x, Q[i].y, Q[i].id, Q[i].f);if(!Q[i].f) tree_add(Q[i].y);else {int t = tree_query(Q[i].y);//printf("%d %d ", Q[i].x, Q[i].y); printf("%d\n", t);tree[Q[i].id][Q[i].f] = t;}}}void doit(void){sort(st+1, st+tot+1);for (int i=1; i<=N; i++){y[i] = got(y[i]);Q[++ all].x = x[i]; Q[all].y = y[i];}for (int i=1; i<=M; i++){b[i] = got(b[i]); d[i] = got(d[i]);Q[++ all].x = c[i], Q[all].y = d[i], Q[all].id = i, Q[all].f = 1;Q[++ all].x = a[i] , Q[all].y = d[i], Q[all].id = i, Q[all].f = 2;Q[++ all].x = c[i], Q[all].y = b[i] , Q[all].id = i, Q[all].f = 3;Q[++ all].x = a[i] , Q[all].y = b[i] , Q[all].id = i, Q[all].f = 4;}work();for (int i=1; i<=M; i++){int ans = 0; ans = tree[i][1] + tree[i][4] - tree[i][2] - tree[i][3];//printf("%d %d %d %d\n", tree[i][1], tree[i][4], tree[i][2], tree[i][3]);//printf("%d\n", ans);}return;}int main(void){N = read(); M = read();for (int i=1; i<=N; i++) x[i] = read(), y[i] = read(), st[++ tot] = y[i];for (int i=1; i<=M; i++)   a[i] = read(), b[i] = read(), c[i] = read(), d[i] = read(),  st[++ tot] = b[i], st[++ tot] = d[i];doit();return 0;}

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