标签:cnt return get int sum 题锦 -- 莫队 结构
zoto【来源:2021杭电多校第一场的第10题】
题目:
Problem Description You are given an array fx.
For each i(1<=i<=n) , we use a point (i,fx[i]) in XoY coordinate plane to described it.
You are asked to answer m queries, in each query there will be a rectangle and you need to count how many different y-cooordinate (of the points mentioned above) in the queried rectangle. Input The first line contains an integer T(1<=T<=5) representing the number of test cases.
For each test case , there are two integers n,m(1<=n<=100000,1<=m<=100000) in the first line.
Then one line contains n integers fx[i](0<=fx[i]<=100000)
Each of the next m lines contain four integers x0,y0,x1,y1(1<=x0<=x1<=n,0<=y0<=y1<=100000) which means matrix's lower-leftmost cell is (x0,y0) and upper-rightest cell is (x1,y1). Output For each test case print a single integer in a new line. Sample Input 1 4 2 1 0 3 1 1 0 4 3 1 0 4 2 Sample Output 3 2 Source 2021“MINIEYE杯”中国大学生算法设计超级联赛(1)
分析:
> 题目:
> T组测试数据, T大小为5:
> 给定n个坐标,横坐标为从1到n,对应纵坐标为fx[i],共有m此询问,其中n,m,fx大小都是1e5
> 每次询问,查询一个以(x0,y0)为左下角,以(x1,y1)为右上角的矩形面积内,有多少个不同的fx
> 解决:
> T是T,m是1e5,那么,我们每次的查询操作的时间复杂度需要降到O(logn)才能合法,而且,只有查询没有修改,基础莫队便可以解决次题。
代码:
#include <bits/stdc++.h> using namespace std; const int N = 100005; int n, m, len; int w[N], cnt[N], sum[N], ans[N]; struct Query{ int l, r, L, R, id; }q[N]; int get(int x) { return x / len; } bool cmp(const Query& a, const Query& b) { int x = get(a.l), y = get(b.l); if (x != y) return x < y; return a.r < b.r; // if (x == y) return a.r < b.r; // return (x & 1) ? x > y : x < y; } void add(int x) { // printf("del: x = %d, cnt[x] = %d\n", x, cnt[x]); //sum计算块内的节点数量,我们对y轴分块,计算的是fx的分块内的值 if (!cnt[x]) sum[get(x)] ++; cnt[x] ++; } void del(int x) { // printf("del: x = %d, cnt[x] = %d\n", x, cnt[x]); cnt[x] --; if (!cnt[x]) sum[get(x)] --; } int calc(int y) { //计算y轴,从0到y的个数 int res = 0; //sum代表y轴分块val的数量,res先求的是前缀和 for (int i = 0; i < get(y); i ++ ) res += sum[i]; //最上面的一点,这些不再整块内,暴力计算 for (int i = get(y) * len; i <= y; i ++ ) res += (cnt[i] >= 1); return res; } void solve() { memset(cnt, 0, sizeof cnt); memset(sum, 0, sizeof sum); scanf("%d%d", &n, &m); // len = sqrt(n); len = 131; for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]); for (int i = 0; i < m; i ++ ) { int l, L, r, R; scanf("%d%d%d%d", &l, &L, &r, &R); q[i] = {l, r, L, R, i}; } sort(q, q + m, cmp); for (int i = 0, l = 1, r = 0; i < m; i ++ ) { // printf("q[i].l = %d, bottom = %d, q[i].r = %d, top = %d\n", q[i].l, q[i].L, q[i].r, q[i].R); // printf("del l ++ \n"); while (l < q[i].l) del(w[l ++ ]); // printf("add -- l \n"); while (l > q[i].l) add(w[-- l ]); // printf("add ++ r \n"); while (r < q[i].r) add(w[++ r ]); // printf("del r -- \n"); while (r > q[i].r) del(w[r -- ]); ans[q[i].id] = calc(q[i].R) - calc(q[i].L - 1); } for (int i = 0; i < m; i ++ ) printf("%d\n", ans[i]); } int main() { int T = 1; cin >> T; while (T -- ) { solve(); } return 0; }
标签:cnt,return,get,int,sum,题锦,--,莫队,结构 来源: https://www.cnblogs.com/Iamcookieandyou/p/15057252.html
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