2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it
链接:https://www.nowcoder.com/acm/contest/163/F
来源:牛客网
2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it
时间限制:C/C++ 3秒,其他语言6秒
空间限制:C/C++ 262144K,其他语言524288K
64bit IO Format: %lld
空间限制:C/C++ 262144K,其他语言524288K
64bit IO Format: %lld
题目描述
There is a matrix A that has N rows and M columns. Each grid (i,j)(0 ≤ i < N, 0 ≤ j < M) is painted in white at first.
Then we perform q operations:
For each operation, we are given (xc, yc) and r. We will paint all grids (i, j) that meets to black.
You need to calculate the number of white grids left in matrix A.
Then we perform q operations:
For each operation, we are given (xc, yc) and r. We will paint all grids (i, j) that meets to black.
You need to calculate the number of white grids left in matrix A.
输入描述:
The first line of the input is T(1≤ T ≤ 40), which stands for the number of test cases you need to solve.
The first line of each case contains three integers N, M and q (1 ≤ N, M ≤ 2 x 104; 1 ≤ q ≤ 200), as mentioned above.
The next q lines, each lines contains three integers xc
, yc
and r (0 ≤ xc
< N; 0 ≤ yc
< M; 0 ≤ r ≤ 105
), as mentioned above.
输出描述:
For each test case, output one number.
示例1
输出
复制729 0
题意还是比较简单的,给你n和m的格子让你去画圆,如果这个点在圆内,就要被染色,问你还剩多少点没有被染色
这个题目可以直接暴力扫描线,也就成了维护线段的并
#include<bits/stdc++.h> using namespace std; const int N=2e5+5; vector<pair<int,int> >V[N]; int main() { int T; scanf("%d",&T); while(T--) { for(int i=0; i<N; i++)V[i].clear(); int n,m,q; scanf("%d%d%d",&n,&m,&q); for(int j=0,x,y,r; j<q; j++) { scanf("%d%d%d",&x,&y,&r); for(int i=max(0,y-r),d; i<=min(m-1,y+r); i++) d=(sqrt(r*r-(y-i)*(y-i)+1e-6)),V[i].push_back(make_pair(max(0,x-d),min(x+d,n-1))); } int s=n*m; for(int i=0; i<m; i++) { int l=V[i].size(); if(l>0) { sort(V[i].begin(),V[i].end()); int r=-1; for(auto X:V[i]) { if(X.second<=r)continue; if(X.first>r) s-=X.second-X.first+1; else s-=X.second-r; r=X.second; } } } cout<<s<<"\n"; } return 0; }
本文来自博客园,作者:暴力都不会的蒟蒻,转载请注明原文链接:https://www.cnblogs.com/BobHuang/p/9493577.html