使用DFS枚举所有的安排。每次加入后,当一个圆。他的立场是最大的,并已加入了圆环中的所有切线位置前面。
14383635 | 10012 | How Big Is It? | Accepted | C++ | 0.086 | 2014-10-20 11:07:33 |
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef double Dou;
#define FOR(i,n) for(int i = 0; i < n; i ++)
const int maxn = 10;
int n,vis[maxn];
Dou ans,r[maxn];
struct POS{ //圆心位置就是(x,r)
Dou x,r; //横坐标,半径
}pos[maxn];
Dou check(int cur,int k){
Dou ret = r[k];
FOR(i,cur){
Dou a = pos[i].r + r[k];
Dou b = pos[i].r - r[k];
Dou c = sqrt(a * a - b * b);
ret = max(ret,pos[i].x + c);
}
pos[cur].x = ret; pos[cur].r = r[k];
return ret + r[k];
}
void dfs(int cur,double L){
if(ans != -1 && L > ans) return;
if(cur == n){
Dou ret;
for(int i = 0; i < cur; i++){
if(i == 0) ret = pos[i].x + pos[i].r;
else ret = max(ret,pos[i].x + pos[i].r);
}
ans = (ans == -1) ?
ret : min(ans,ret); return; } FOR(i,n) if(!vis[i]){ vis[i] = 1; dfs(cur + 1,check(cur,i)); vis[i] = 0; } return; } int main(){ int T; scanf("%d",&T); while(T--){ scanf("%d",&n); ans = -1; memset(vis,0,sizeof(vis)); FOR(i,n) scanf("%lf",&r[i]); dfs(0,0.0); printf("%.3f\n",ans); } return 0; }