luogu1220 关路灯

题目大意

　　路面上有一些开着的灯，每个灯有功率和它的位置，人在第c个灯处，行走速度1m/s。问怎样关灯能使耗能最小，输出这个耗能。

思路

　　

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_N = 60, INF = 0x3f3f3f3f;
int F[MAX_N][MAX_N][2], Pos[MAX_N], W[MAX_N];
int WPrefix[MAX_N];
int N, Start;

void GetWPrefix()
{
	for (int i = 1; i <= N; i++)
		WPrefix[i] = WPrefix[i - 1] + W[i];
}

int DP()
{
	memset(F, INF, sizeof(F));
	F[Start][Start][0] = F[Start][Start][1] = 0;
	for (int len = 2; len <= N; len++)
		for (int i = 1; i <= N - len + 1; i++)
		{
			int j = i + len - 1;
			F[i][j][1] = min(F[i][j][1], F[i][j - 1][1] + (Pos[j] - Pos[j - 1]) * (WPrefix[i - 1] + WPrefix[N] - WPrefix[j - 1]));
			F[i][j][1] = min(F[i][j][1], F[i][j - 1][0] + (Pos[j] - Pos[i]) * (WPrefix[i - 1] + WPrefix[N] - WPrefix[j - 1]));
			F[i][j][0] = min(F[i][j][0], F[i + 1][j][0] + (Pos[i + 1] - Pos[i]) * (WPrefix[i] + WPrefix[N] - WPrefix[j]));
			F[i][j][0] = min(F[i][j][0], F[i + 1][j][1] + (Pos[j] - Pos[i]) * (WPrefix[i] + WPrefix[N] - WPrefix[j]));
		}
	return min(F[1][N][0], F[1][N][1]);
}

int main()
{
	scanf("%d%d", &N, &Start);
	for (int i = 1; i <= N; i++)
		scanf("%d%d", Pos + i, W + i);
	GetWPrefix();
	printf("%d
", DP());
}