传送门

如果去关某一个灯，那么途中经过的灯都能关闭，那么就是连续一段区间，区间DP。

f[i][j][0] 表示关完 i, j 这个区间且在 i 这个位置

f[i][j][1] 表示关完 i, j 这个区间且在 j 这个位置

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 1001
#define min(x, y) ((x) < (y) ? (x) : (y))

int n, s;
int f[N][N][2], a[N], d[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 << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

int main()
{
	int i, j;
	n = read();
	s = read();
	for(i = 1; i <= n; i++) d[i] = read(), a[i] = read() + a[i - 1];
	memset(f, 127 / 3, sizeof(f));
	f[s][s][0] = f[s][s][1] = 0;
	for(i = s; i >= 1; i--)
		for(j = i + 1; j <= n; j++)
		{
			f[i][j][0] = min(f[i][j][0], f[i + 1][j][0] + (d[i + 1] - d[i]) * (a[n] - (a[j] - a[i])));
			f[i][j][0] = min(f[i][j][0], f[i + 1][j][1] + (d[j] - d[i]) * (a[n] - (a[j] - a[i])));
			f[i][j][1] = min(f[i][j][1], f[i][j - 1][0] + (d[j] - d[i]) * (a[n] - (a[j - 1] - a[i - 1])));
			f[i][j][1] = min(f[i][j][1], f[i][j - 1][1] + (d[j] - d[j - 1]) * (a[n] - (a[j - 1] - a[i - 1])));
		}
	printf("%d\n", min(f[1][n][0], f[1][n][1]));
	return 0;
}