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School Personal Contest #1 (Winter Computer School 2010/11) - Codeforces Beta Round #38 (ACM-ICPC Ru
作者:互联网
比赛链接:
https://codeforces.com/contest/38
D. Vasya the Architect
题目大意:
有 \(n\) 个立方体,给了每一个立方体在 \(xoy\) 面上投影的对角坐标,按照给定的顺序,依次往上叠立方体,问在保持平衡不到的状态下,最多能叠多少个立方体。
思路:
对于第 \(i\) 个立方体,需要判断的是它的重心是否落在下面立方体的投影内,即求重心公式,判断是否能保持平衡。
\(k\) 个物品的叠加后的重心落在 \(( (x[1] * w[1] + x[2] * w[2] + ... + x[k] * w[k]) / (w[1] + w[2] + ... + w[k]), (y[1] * w[1] + y[2] * w[2] + ... + y[k] * w[k]) / (w[1] + w[2] + ... + w[k]))\)。
\(w[i]\) 为第 \(i\) 个物品的重量,\(x[i]\) 为第 \(i\) 个物品重心的横坐标,\(y[i]\) 为第 \(i\) 个物品重心的纵坐标。
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
struct node{
double x1, y1, x2, y2, x, y, w;
}a[N];
int n;
void solve(){
for (int i = 0; i < n; i ++ ){
double x = a[i].x * a[i].w, y = a[i].y * a[i].w, w = a[i].w;
for (int j = i - 1; j >= 0; j -- ){
if (x / w < a[j].x1 || x / w > a[j].x2 || y / w < a[j].y1 || y / w > a[j].y2){
cout << i << "\n";
return;
}
x += a[j].x * a[j].w;
y += a[j].y * a[j].w;
w += a[j].w;
}
}
cout << n << "\n";
}
int main(){
cin >> n;
for (int i = 0; i < n; i ++ ){
cin >> a[i].x1 >> a[i].y1 >> a[i].x2 >> a[i].y2;
a[i].x = (a[i].x1 + a[i].x2) / 2.0;
a[i].y = (a[i].y1 + a[i].y2) / 2.0;
if (a[i].x1 > a[i].x2) swap(a[i].x1, a[i].x2);
if (a[i].y1 > a[i].y2) swap(a[i].y1, a[i].y2);
double t = a[i].y2 - a[i].y1;
a[i].w = t * t * t;
}
solve();
return 0;
}
标签:11,School,38,int,x1,立方体,x2,y1,y2 来源: https://www.cnblogs.com/Hamine/p/16178130.html