题意:中文题。
析:按x排序,然后用树状数组维护 y 即可。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<LL, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-3; const int maxn = 500000 + 10; const int maxm = 1e7 + 10; const int mod = 1000000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, -1, 0, 1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } struct Query{ int id, x, y, ty, w; Query() { } Query(int i, int xx, int yy, int t, int ww) : id(i), x(xx), y(yy), ty(t), w(ww) { } bool operator < (const Query &q) const{ return x < q.x || x == q.x && ty < q.ty; } }; Query q[maxn*5], tmp[maxn*5]; int ans[maxn]; int sum[maxm]; inline int lowbit(int x){ return -x&x; } void add(int x, int c){ while(x < maxm){ sum[x] += c; x += lowbit(x); } } int query(int x){ int ans = 0; while(x){ ans += sum[x]; x -= lowbit(x); } return ans; } void dfs(int l, int r){ if(l == r) return ; int m = l + r >> 1; dfs(l, m); dfs(m+1, r); int i = l, j = m+1, k = l; while(i <= m && j <= r){ if(q[i] < q[j]){ tmp[k++] = q[i]; if(!q[i].ty) add(q[i].y, 1); ++i; } else{ tmp[k++] = q[j]; if(q[j].ty) ans[q[j].id] += query(q[j].y) * q[j].w; ++j; } } while(i <= m){ tmp[k++] = q[i]; if(!q[i].ty) add(q[i].y, 1); ++i; } while(j <= r){ tmp[k++] = q[j]; if(q[j].ty) ans[q[j].id] += query(q[j].y) * q[j].w; ++j; } for(int i = l; i <= r; ++i){ if(!q[i].ty && i <= m) add(q[i].y, -1); q[i] = tmp[i]; } } int main(){ scanf("%d %d", &n, &m); int cnt = 0; for(int i = 0; i < n; ++i){ int x, y; scanf("%d %d", &x, &y); q[cnt++] = Query(-1, x+1, y+1, 0, 0); } for(int i = 0; i < m; ++i){ int x1, y1, x2, y2; scanf("%d %d %d %d", &x1, &y1, &x2, &y2); q[cnt++] = Query(i, x1, y1, 1, 1); q[cnt++] = Query(i, x2+1, y2+1, 1, 1); q[cnt++] = Query(i, x1, y2+1, 1, -1); q[cnt++] = Query(i, x2+1, y1, 1, -1); } dfs(0, cnt-1); for(int i = 0; i < m; ++i) printf("%d ", ans[i]); return 0; }