2037: [Sdoi2008]Sue的小球

2037: [Sdoi2008]Sue的小球

链接

题解

　　论文

代码

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>

 using namespace std;

 const int N = ;

 struct Node{
     int x,y,v;
     bool operator < (const Node &a) const {
         return x < a.x;
     }
 }d[N];
 int dp[N][N][],sum[N];

 inline int read() {
     int x = ,f = ;char ch=getchar();
     for (; !isdigit(ch); ch=getchar()) if(ch=='-')f=-;
     for (; isdigit(ch); ch=getchar()) x=x*+ch-'';
     return x*f;
 }

 int main() {
     int n = read(), x0 = read();
     for (int i=; i<=n; ++i) d[i].x = read();
     for (int i=; i<=n; ++i) d[i].y = read();
     for (int i=; i<=n; ++i) d[i].v = read();

     sort(d+,d+n+);
     for (int i=; i<=n; ++i) sum[i] = sum[i-] + d[i].v; // 对下降速度求前缀和 

     for (int i=; i<=n; ++i)
         dp[i][i][] = dp[i][i][] = d[i].y - abs(x0-d[i].x)*sum[n]; //表示已经收集了[i,i]这个区间 

     for (int k=; k<=n; ++k) {
         for (int i=; i+k-<=n; ++i) {
             int j = i + k - ;
             dp[i][j][] = max(dp[i+][j][] + d[i].y - (sum[n] - sum[j] + sum[i]) * (d[i+].x - d[i].x),
                               dp[i+][j][] + d[i].y - (sum[n] - sum[j] + sum[i]) * (d[j].x - d[i].x));
             dp[i][j][] = max(dp[i][j-][] + d[j].y - (sum[n] - sum[j-] + sum[i-]) * (d[j].x - d[i].x),
                               dp[i][j-][] + d[j].y - (sum[n] - sum[j-] + sum[i-]) * (d[j].x - d[j-].x));
         }
     }

     double ans = 1.0*(max(dp[][n][],dp[][n][]))/1000.0;
     printf("%.3lf",ans);

     return ;
 }