洛谷1378 油滴擴展 dfs進行回溯搜索

題目連接：https://www.luogu.com.cn/problem/P1378

題目中給出矩形的長寬和一些點，能夠在每一個點放油滴，油滴會擴展，直到觸碰到矩形的周邊或者其餘油滴的邊緣，求出剩餘面積的最小值，就是求油滴面積的最大值。策略是dfs加上回溯，暴力求解。

代碼以下：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef unsigned int ui;
 4 typedef long long ll;
 5 typedef unsigned long long ull;
 6 #define pf printf
 7 #define mem(a,b) memset(a,b,sizeof(a))
 8 #define prime1 1e9+7
 9 #define pi 3.14159265
10 #define prime2 1e9+9
11 #define scand(x) scanf("%lf",&x) 
12 #define f(i,a,b) for(int i=a;i<=b;i++)
13 #define scan(a) scanf("%d",&a)
14 #define dbg(args) cout<<#args<<":"<<args<<endl;
15 #define pb(i) push_back(i)
16 #define ppb(x) pop_back(x)
17 #define inf 0x3f3f3f3f
18 #define maxn 10
19 int n;
20 double x1,x2,Y1,y2;//瞄的，y1洛谷編譯不了 
21 bool vis[maxn];
22 double dis[maxn][maxn],area=0.0;
23 struct p{
24     double x,y;
25     double r;
26 }cir[maxn];
27 double distance(double x1,double Y1,double x2,double y2)//點距 
28 {
29     return sqrt((x1-x2)*(x1-x2)+(Y1-y2)*(Y1-y2));
30 }
31 double get_max(int num)//計算第num個點造成圓的最大半徑 
32 {
33     double tmp1=min(fabs(cir[num].x-x1),fabs(x2-cir[num].x));
34     double tmp2=min(fabs(cir[num].y-Y1),fabs(y2-cir[num].y));
35     double ans=min(tmp1,tmp2);
36     f(i,1,n)
37     {
38         if((i!=num)&&vis[i])
39         {
40             if(dis[i][num]<=cir[i].r)return 0.0;//一旦被包圍在內就不能夠擴散 
41             ans=min(ans,dis[i][num]-cir[i].r);
42         }
43     }
44     return ans;
45 }
46 void dfs(int dep,double tot)
47 {
48     if(dep==n)
49     {
50         //dbg(area);
51         area=max(area,tot);
52         return;
53     }
54     f(i,1,n)
55     {
56         if(!vis[i])
57         {
58             vis[i]=true;
59             double R=get_max(i);
60             cir[i].r=R;
61             dfs(dep+1,tot+pi*R*R);
62             vis[i]=false;
63             cir[i].r=0.0;//回溯 
64         }
65         
66     }
67 }
68 int main()
69 {
70     //freopen("input.txt","r",stdin);
71     //freopen("output.txt","w",stdout);
72     std::ios::sync_with_stdio(false);
73     scan(n);
74     scanf("%lf %lf %lf %lf",&x1,&Y1,&x2,&y2);
75     f(i,1,n)
76     {
77         scand(cir[i].x);
78         scand(cir[i].y);
79     }
80     mem(vis,false);
81     f(i,1,n)
82         f(j,1,n)
83         {
84             dis[i][j]=distance(cir[i].x,cir[i].y,cir[j].x,cir[j].y);
85         }
86     dfs(0,0.0);
87     int ans=(fabs(x1-x2)*fabs(Y1-y2)-area+0.5);
88     pf("%d\n",ans);    
89  }