洛谷1378 油滴扩展 dfs进行回溯搜索

题目链接：https://www.luogu.com.cn/problem/P1378

题目中给出矩形的长宽和一些点，可以在每个点放油滴，油滴会扩展，直到触碰到矩形的周边或者其他油滴的边缘，求出剩余面积的最小值，就是求油滴面积的最大值。策略是dfs加上回溯，暴力求解。

代码如下：

 #include<bits/stdc++.h>
 using namespace std;
 typedef unsigned int ui;
 typedef long long ll;
 typedef unsigned long long ull;
 #define pf printf
 #define mem(a,b) memset(a,b,sizeof(a))
 #define prime1 1e9+7
 #define pi 3.14159265
 #define prime2 1e9+9
 #define scand(x) scanf("%lf",&x)
 #define f(i,a,b) for(int i=a;i<=b;i++)
 #define scan(a) scanf("%d",&a)
 #define dbg(args) cout<<#args<<":"<<args<<endl;
 #define pb(i) push_back(i)
 #define ppb(x) pop_back(x)
 #define inf 0x3f3f3f3f
 #define maxn 10
 int n;
 double x1,x2,Y1,y2;//瞄的，y1洛谷编译不了
 bool vis[maxn];
 double dis[maxn][maxn],area=0.0;
 struct p{
     double x,y;
     double r;
 }cir[maxn];
 double distance(double x1,double Y1,double x2,double y2)//点距
 {
     return sqrt((x1-x2)*(x1-x2)+(Y1-y2)*(Y1-y2));
 }
 double get_max(int num)//计算第num个点形成圆的最大半径
 {
     double tmp1=min(fabs(cir[num].x-x1),fabs(x2-cir[num].x));
     double tmp2=min(fabs(cir[num].y-Y1),fabs(y2-cir[num].y));
     double ans=min(tmp1,tmp2);
     f(i,,n)
     {
         if((i!=num)&&vis[i])
         {
             if(dis[i][num]<=cir[i].r)return 0.0;//一旦被包围在内就不可以扩散
             ans=min(ans,dis[i][num]-cir[i].r);
         }
     }
     return ans;
 }
 void dfs(int dep,double tot)
 {
     if(dep==n)
     {
         //dbg(area);
         area=max(area,tot);
         return;
     }
     f(i,,n)
     {
         if(!vis[i])
         {
             vis[i]=true;
             double R=get_max(i);
             cir[i].r=R;
             dfs(dep+,tot+pi*R*R);
             vis[i]=false;
             cir[i].r=0.0;//回溯
         }

     }
 }
 int main()
 {
     //freopen("input.txt","r",stdin);
     //freopen("output.txt","w",stdout);
     std::ios::sync_with_stdio(false);
     scan(n);
     scanf("%lf %lf %lf %lf",&x1,&Y1,&x2,&y2);
     f(i,,n)
     {
         scand(cir[i].x);
         scand(cir[i].y);
     }
     mem(vis,false);
     f(i,,n)
         f(j,,n)
         {
             dis[i][j]=distance(cir[i].x,cir[i].y,cir[j].x,cir[j].y);
         }
     dfs(,0.0);
     int ans=(fabs(x1-x2)*fabs(Y1-y2)-area+0.5);
     pf("%d\n",ans);
  }