[luoguP1103] 书本整理（DP）

传送门

以 去掉多少个 为阶段不好做。

去掉 k 个也可以变成选 n - k 个

f[i][j] 表示前 i 个数中 选 j 个的最优解，a[i] 必选

f[i][j] = min(f[i][j], f[k][j - 1] + abs(b[k] - b[i])) (2 <= j <= min(i, n - m), j - 1 <= k < i)

——代码

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

const int MAXN = 101, INF = ~(1 << 31);
int n, m, ans = INF;
int f[MAXN][MAXN];

struct node
{
    int a, b;
}p[MAXN];

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 << 1) + (x << 3) + ch - '0';
    return x * f;
}

inline bool cmp(node x, node y)
{
    return x.a < y.a;
}

inline int min(int x, int y)
{
    return x < y ? x : y;
}

inline int abs(int x)
{
    return x < 0 ? -x : x;
}

int main()
{
    int i, j, k;
    n = read();
    m = n - read();
    for(i = 1; i <= n; i++) p[i].a = read(), p[i].b = read();
    std::sort(p + 1, p + n + 1, cmp);
    for(i = 2; i <= n; i++)
        for(j = 2; j <= min(i, m); j++)
        {
            f[i][j] = INF;
            for(k = j - 1; k < i; k++)
                f[i][j] = min(f[i][j], f[k][j - 1] + abs(p[k].b - p[i].b));
        }
    for(i = m; i <= n; i++) ans = min(ans, f[i][m]);
    printf("%d
", ans);
    return 0;
}