$dp[i][j][0/1]$关掉区间$[i,j]$最后一步关的是$i/j$的最小花费

要转移我们就要考虑上一次是在哪里更新来的，这一次关掉的$i,j$的花费是多少，我们再开一个数组$g[i][j][0/1]$表示关掉区间$[i,j]$最后一步关$i/j$的最优方案花的时间

然后转移的时候优先选花费小的，花费一样就选时间短的，留给后面的机会大

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=60;
int c,w[maxn],n,dis[maxn],dp[maxn][maxn][2],g[maxn][maxn][2];
int dist(int x,int y){
	return abs(dis[x]-dis[y]);
}
int main(){
	memset(dp,0x3f,sizeof(dp));
	scanf("%d%d",&n,&c);
	for(int i=1;i<=n;i++) scanf("%d%d",&dis[i],&w[i]);
	for(int i=1;i<=n;i++) dp[i][i][0]=dp[i][i][1]=dist(i,c)*w[i],g[i][i][0]=g[i][i][1]=dist(i,c); 
	for(int d=2;d<=n;d++)	
		for(int i=1;i<=n-d+1;i++){
			int j=i+d-1;
			if(dp[i+1][j][0]+(dist(i+1,i)+g[i+1][j][0])*w[i]<dp[i+1][j][1]+(dist(i,j)+g[i+1][j][1])*w[i])
				g[i][j][0]=(dist(i+1,i)+g[i+1][j][0]),dp[i][j][0]=dp[i+1][j][0]+g[i][j][0]*w[i];
			else if(dp[i+1][j][0]+(dist(i+1,i)+g[i+1][j][0])*w[i]>dp[i+1][j][1]+(dist(i,j)+g[i+1][j][1])*w[i])
				g[i][j][0]=(dist(i,j)+g[i+1][j][1]),dp[i][j][0]=dp[i+1][j][1]+g[i][j][0]*w[i];
			else if(dist(i+1,i)+g[i+1][j][0]<dist(i,j)+g[i+1][j][1])
					g[i][j][0]=(dist(i+1,i)+g[i+1][j][0]),dp[i][j][0]=dp[i+1][j][0]+g[i][j][0]*w[i];
			else g[i][j][0]=(dist(i,j)+g[i+1][j][1]),dp[i][j][0]=dp[i+1][j][1]+g[i][j][0]*w[i];
			if(dp[i][j-1][1]+(dist(j-1,j)+g[i][j-1][1])*w[j]<dp[i][j-1][0]+(dist(i,j)+g[i][j-1][0])*w[j])
				g[i][j][1]=(dist(j-1,j)+g[i][j-1][1]),dp[i][j][1]=dp[i][j-1][1]+g[i][j][1]*w[j];
			else if(dp[i][j-1][1]+(dist(j-1,j)+g[i][j-1][1])*w[j]>dp[i][j-1][0]+(dist(i,j)+g[i][j-1][0])*w[j])
				g[i][j][1]=(dist(i,j)+g[i][j-1][0]),dp[i][j][1]=dp[i][j-1][0]+g[i][j][1]*w[j];
			else if(dist(j-1,j)+g[i][j-1][1]<dist(i,j)+g[i][j-1][0])
				g[i][j][1]=(dist(j-1,j)+g[i][j-1][1]),dp[i][j][1]=dp[i][j-1][1]+g[i][j][1]*w[j];
			else g[i][j][1]=(dist(i,j)+g[i][j-1][0]),dp[i][j][1]=dp[i][j-1][0]+g[i][j][1]*w[j];
		}
	printf("%d",min(dp[1][n][0],dp[1][n][1]));
	return 0;
} 

但是70pts WA#6#7#10求神犇调如果思路错了求个简单一点的反例