题目意思就是说,给你一张star maps,每个star有自己的level,level是这样计算的:(Let the level of a star be an amount of the stars that are not higher and not to the right
of the given star.)统计这个star左下角star的个数,就是这个star的level。现在要你总计图中level从0到N-1的star分别有多少个。
如入是按照X Y Y的递增顺序输入的 Y相同时按照X的递增顺序 所以只要求X之前小于等于X的星星的个数就可以了 也就是按照X的值建树
#include <iostream> #include<cstdio> #include<cstring> using namespace std; #define maxn 32005 int tree[maxn]; int n; int ans[15005]; inline int Lowbit(int x) { return x & (-x); } inline void Update(int pos, int val) { while (pos <= maxn) { tree[pos] += val; pos += Lowbit(pos); } } inline int Query(int x) { int sum = 0; while (x > 0) { sum += tree[x]; x -= Lowbit(x); } return sum; } int main() { int x,y; while(~scanf("%d",&n)) { memset(tree,0,sizeof(tree)); memset(ans,0,sizeof(ans)); for(int i=0; i<n; i++) { scanf("%d%d",&x,&y); ans[Query(x+1)]++; Update(x+1,1); } for(int i=0; i<n; i++) printf("%d\n",ans[i]); } return 0; }
时间: 2024-10-17 09:47:29