[luogu2446][bzoj2037][SDOI2008] Sue的小球【区间DP】

分析

简单区间DP，
定义状态f[i][j][0/1]为取完i-j的小球最后取i/j上的小球所能获得的最大价值。
排序转移。

ac代码

#include <bits/stdc++.h>
#define N 1005
using namespace std;
template <typename T>
inline void read(T &x) {
    x = 0; T fl = 1; char ch = 0;
    for (; ch < '0' || ch > '9'; ch = getchar())
        if (ch == '-') fl = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        x = (x << 1) + (x << 3) + (ch ^ 48);
    x *= fl;
}
struct node {
    int x, y, val;
}a[N];
int f[N][N][2], sum[N];
int n, x0;
bool cmp(const node &a, const node &b) {
    return a.x == b.x? a.y < b.y: a.x < b.x;
}
int main() {
    read(n); read(x0);
    for (int i = 1; i <= n; i ++) read(a[i].x);
    for (int i = 1; i <= n; i ++) read(a[i].y);
    for (int i = 1; i <= n; i ++) read(a[i].val);
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + a[i].val;
    for (int i = 1; i <= n; i ++) 
        f[i][i][0] = f[i][i][1] = a[i].y - abs(a[i].x - x0) * sum[n];
    for (int i = 2; i <= n; i ++) {
        for (int j = 1; j <= n - i + 1; j ++) {
            int k = j + i - 1;
            f[j][k][0] = max(f[j + 1][k][1] + a[j].y - (sum[n] - sum[k] + sum[j]) * abs(a[k].x - a[j].x), f[j + 1][k][0] + a[j].y - (sum[n] - sum[k] + sum[j]) * abs(a[j + 1].x - a[j].x));
            f[j][k][1] = max(f[j][k - 1][0] + a[k].y - (sum[n] - sum[k - 1] + sum[j - 1]) * abs(a[k].x - a[j].x), f[j][k - 1][1] + a[k].y - (sum[n] - sum[k - 1] + sum[j - 1]) * abs(a[k].x - a[k - 1].x));
        }
    }
    printf("%.3lf\n", 1.0 * max(f[1][n][1], f[1][n][0]) / 1000.0);
    return 0;
}