poj1819Disks

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题意：从左到右按顺序给你n个圆的半径，把左右两边想象成两堵墙的话，就是左右两边向里挤压，问哪些圆是对最后的宽度不影响。

刚开始理解错了，。。以为怎么放圆使宽度最小。。

这样就可以尽量使每个圆向左靠，找出当前圆与最左相切的圆，他们之间的那些圆肯定就是可以消除的，特判一下最左和最右就可以了。

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 1010

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 struct point

18 {

19     double x,y,r;

20     point(double x=0,double y=0,double r =0):x(x),y(y),r(r){}

21 }p[N];

22 int o[N];

23 bool f[N];

24 typedef point pointt;

25 pointt operator -(point a,point b)

26 {

27     return point(a.x-b.x,a.y-b.y);

28 }

29 int dcmp(double x)

30 {

31     if(fabs(x)<eps) return 0;

32     return x<0?-1:1;

33 }

34 double dis(point a)

35 {

36     return sqrt(a.x*a.x+a.y*a.y);

37 }

38 int main()

39 {

40     int n,i,j;

41     while(scanf("%d",&n)!=EOF)

42     {

43         memset(f,0,sizeof(f));

44         for(i = 1; i <=  n; i++)

45         scanf("%lf",&p[i].r);

46         p[1] = point(p[1].r,0,p[1].r);

47         p[0].r = p[0].x = 0;

48         double rig;

49         for(i = 2; i <= n; i++)

50         {

51             double maxz = 0;

52             int x = i-1;

53             for(j = i-1 ; j >= 1 ; j--)

54             {

55                 double d = sqrt((p[j].r+p[i].r)*(p[j].r+p[i].r)-(p[j].r-p[i].r)*(p[j].r-p[i].r))+p[j].x;

56                 if(dcmp(d-maxz)>=0)

57                 {

58                     maxz = max(maxz,d);

59                     x = j;

60                 }

61             }

62             if(dcmp(maxz-p[i].r)<=0)

63             {

64                 p[i].x = p[i].r;

65                 for(j = 1 ; j < i; j++)

66                 f[j] = 1;

67                 continue;

68             }

69             p[i].x = maxz;

70             //cout<<i<<" "<<x<<" "<<maxz<<endl;

71             for(j = x+1 ; j < i; j++)

72             f[j] = 1;

73         }

74         int x = n;

75         for(i = 1; i <= n ;i ++)

76         {

77             if(dcmp(p[i].r+p[i].x-rig)>0)

78             {

79                 rig = max(rig,p[i].r+p[i].x);

80                 x = i;

81             }

82         }

83         for(i = x+1 ; i <= n ;i++)

84         f[i] = 1;

85         int g = 0;

86         for(i = 1; i <= n ;i++)

87         if(f[i])

88         o[++g] = i;

89         cout<<g<<endl;

90         for(i = 1; i <= g ;i++)

91         cout<<o[i]<<endl;

92     }

93     return 0;

94 }

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