[USACO08FEB]修路Making the Grade

题目链接：

走这里

题目分析：

考虑绝对值的几何意义，显然(b)里的数一定在(a)里出现过
离不离散化问题不大，用下标作第二位状态就行
设(dp[i][j])表示第(i)个数，高度为(a[j])时的最优解
方程见代码

代码：

#include<bits/stdc++.h>
#define int long long
#define N (2000 + 10)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
	return cnt * f;
}
int n, a[N], b[N << 1], dp[N][N], gmin = 1 << 30;
int ans = (1 << 30);
signed main() {
//	freopen("grading.in", "r", stdin);
//	freopen("grading.out", "w", stdout);
	n = read();
	for (register int i = 1; i <= n; ++i) a[i] = b[i] = read();
	memset(dp, 0x3f, sizeof(dp));
	for (register int i = 1; i <= n; ++i) dp[1][i] = abs(a[1] - a[i]);
	sort(a + 1, a + n + 1);
	
	for (register int i = 2; i <= n; ++i) {
		gmin = (1 << 30);
		for (register int j = 1; j <= n; ++j) {
			gmin = min(gmin, dp[i - 1][j]);
			dp[i][j] = gmin + abs(b[i] - a[j]);
		}
	}
	for (register int i = 1; i <= n; ++i) ans = min(ans, dp[n][i]);

	memset(dp, 0x3f, sizeof(dp));
	
	for (register int i = 2; i <= n; ++i) {
		gmin = (1 << 30);
		for (register int j = n; j >= 1; --j) {
			gmin = min(gmin, dp[i - 1][j]);
			dp[i][j] = min(dp[i][j], gmin + abs(b[i] - a[j]));
		}
	}
	
	for (register int i = 1; i <= n; ++i) ans = min(ans, dp[n][i]);
	printf("%lld", ans);
	return 0;
}