[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

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cmath>
 #include <string>
 #include <sstream>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 #include <queue>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <set>
 #include <map>
 typedef long long LL;
 typedef long double LD;
 #define pi acos(-1.0)
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 typedef pair<int, int> PI;
 typedef pair<int, PI> PP;
 #ifdef _WIN32
 #define LLD "%I64d"
 #else
 #define LLD "%lld"
 #endif
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 //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;}
 //inline void print(LL x){printf(LLD, x);puts("");}
 //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();}}

 const double eps=1e-;
 struct node
 {
     LL x, y, t;
     double p;
 }a[];
 double dp[];
 bool cmp(node a, node b)
 {
     return a.t<b.t;
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif
     int n;
     while(~scanf("%d", &n))
     {
         for(int i=;i<n;i++)
         {
             scanf(LLD, &a[i].x);
             scanf(LLD, &a[i].y);
             scanf(LLD, &a[i].t);
             scanf("%lf", &a[i].p);
         }
         sort(a, a+n, cmp);
         memset(dp, , sizeof(dp));
         for(int i=;i<n;i++)
         {
             dp[i]=a[i].p;
             for(int j=;j<i;j++)
                 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)
                     dp[i]=max(dp[i], a[i].p+dp[j]);
         }
         double ans=;
         for(int i=;i<n;i++)
             ans=max(ans, dp[i]);
         printf("%.10lf\n", ans);
     }
     return ;
 }

Codeforces 30C