[dp]Codeforces30C Shooting Gallery
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 }
