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

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

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

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