[dp]Codeforces30C Shooting Gallery

题目链接

题意: 给n个点 每个点的坐标 x y 出现的时间t 射中的概率

从i点到j点的时间为它们的距离.

求射中个数的最大期望

很水的dp  坑点就是要用LL

 1 #include <cstdio>

 2 #include <cstdlib>

 3 #include <cstring>

 4 #include <climits>

 5 #include <cctype>

 6 #include <cmath>

 7 #include <string>

 8 #include <sstream>

 9 #include <iostream>

10 #include <algorithm>

11 #include <iomanip>

12 using namespace std;

13 #include <queue>

14 #include <stack>

15 #include <vector>

16 #include <deque>

17 #include <set>

18 #include <map>

19 typedef long long LL;

20 typedef long double LD;

21 #define pi acos(-1.0)

22 #define lson l, m, rt<<1

23 #define rson m+1, r, rt<<1|1

24 typedef pair<int, int> PI;

25 typedef pair<int, PI> PP;

26 #ifdef _WIN32

27 #define LLD "%I64d"

28 #else

29 #define LLD "%lld"

30 #endif

31 //#pragma comment(linker, "/STACK:1024000000,1024000000")

32 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}

33 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}

34 //inline void print(LL x){printf(LLD, x);puts("");}

35 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}

36 

37 const double eps=1e-6;

38 struct node

39 {

40     LL x, y, t;

41     double p;

42 }a[1005];

43 double dp[100005];

44 bool cmp(node a, node b)

45 {

46     return a.t<b.t;

47 }

48 int main()

49 {

50 #ifndef ONLINE_JUDGE

51     freopen("in.txt", "r", stdin);

52     freopen("out.txt", "w", stdout);

53 #endif

54     int n;

55     while(~scanf("%d", &n))

56     {

57         for(int i=0;i<n;i++)

58         {

59             scanf(LLD, &a[i].x);

60             scanf(LLD, &a[i].y);

61             scanf(LLD, &a[i].t);

62             scanf("%lf", &a[i].p);

63         }

64         sort(a, a+n, cmp);

65         memset(dp, 0, sizeof(dp));

66         for(int i=0;i<n;i++)

67         {

68             dp[i]=a[i].p;

69             for(int j=0;j<i;j++)

70                 if((a[j].x-a[i].x)*(a[j].x-a[i].x)+(a[j].y-a[i].y)*(a[j].y-a[i].y)<=(a[j].t-a[i].t)*(a[j].t-a[i].t)+eps)

71                     dp[i]=max(dp[i], a[i].p+dp[j]);

72         }

73         double ans=0;

74         for(int i=0;i<n;i++)

75             ans=max(ans, dp[i]);

76         printf("%.10lf\n", ans);

77     }

78     return 0;

79 }

Codeforces 30C