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 = 1010;

struct Node{
    int x,y,v;
    bool operator < (const Node &a) const {
        return x < a.x;
    }
}d[N];
int dp[N][N][2],sum[N];

inline int read() {
    int x = 0,f = 1;char ch=getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-')f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x*f;
}

int main() {
    int n = read(), x0 = read();
    for (int i=1; i<=n; ++i) d[i].x = read();
    for (int i=1; i<=n; ++i) d[i].y = read();
    for (int i=1; i<=n; ++i) d[i].v = read();
    
    sort(d+1,d+n+1);
    for (int i=1; i<=n; ++i) sum[i] = sum[i-1] + d[i].v; // 对下降速度求前缀和 
    
    for (int i=1; i<=n; ++i) 
        dp[i][i][0] = dp[i][i][1] = 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=2; k<=n; ++k) {
        for (int i=1; i+k-1<=n; ++i) { 
            int j = i + k - 1;
            dp[i][j][0] = max(dp[i+1][j][0] + d[i].y - (sum[n] - sum[j] + sum[i]) * (d[i+1].x - d[i].x),
                              dp[i+1][j][1] + d[i].y - (sum[n] - sum[j] + sum[i]) * (d[j].x - d[i].x));
            dp[i][j][1] = max(dp[i][j-1][0] + d[j].y - (sum[n] - sum[j-1] + sum[i-1]) * (d[j].x - d[i].x),
                              dp[i][j-1][1] + d[j].y - (sum[n] - sum[j-1] + sum[i-1]) * (d[j].x - d[j-1].x));
        }
    }
    
    double ans = 1.0*(max(dp[1][n][0],dp[1][n][1]))/1000.0;
    printf("%.3lf",ans);

    return 0;
}