ACM-ICPC 2015 Asia Tsukuba Regional Online Open Contest B(两挡板之间放着圆球,求挡板最短距离)
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http://judge.u-aizu.ac.jp/onlinejudge/contest_problem.jsp?id=ICPCOOC2015
一开始想得很麻烦,然后发现又读漏了可以简化题目的信息……好不容易想到的方法一定要记录一下。。。
#include<iostream> #include<algorithm> #include<string> #include<map>//ll dx[4]={0,0,-1,1};ll dy[4]={-1,1,0,0}; #include<set>// #include<vector> #include<cmath> #include<stack> #include<string.h> #include<stdlib.h> #include<cstdio> #define mod 1e9+7 #define ll long longusing namespace std; double y[1010];//存每个球的中心的坐标 double x[1010];int main(){int n;cin>>n;for(int i=0;i<n;++i)cin>>x[i];double s=0;if(n==1){printf("%.6f\n",2*x[0]);return 0;}y[0]=x[0];for(int i=1;i<n;++i){double g=sqrt((x[i]+x[i-1])*(x[i]+x[i-1])-(x[i]-x[i-1])*(x[i]-x[i-1]));double p=y[i-1]+g; //先算出这个球对于前一个球的最佳摆放坐标 for(int j=i-2;j>=0;--j){g=sqrt((p-y[j])*(p-y[j])+(x[i]-x[j])*(x[i]-x[j]));if(g<x[i]+x[j]){ //判断球摆放在这里会不会和之前的球有冲突重叠 double h=sqrt((x[i]+x[j])*(x[i]+x[j])-(x[i]-x[j])*(x[i]-x[j]));p=max(p,y[j]+h);}}y[i]=max(p,x[i]); //这里还要注意和自身的半径比较 }double maxx=0;for(int i=0;i<n;++i){maxx=max(maxx,y[i]+x[i]); //【球的坐标+它的半径】最大值就是挡板之间的距离}printf("%.6f\n",maxx);return 0;}
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