/* * poj2374 AC * 线段树+DP 这道题还是很典型的，值得一做。 * * */#include<cstdio>#include<algorithm>#define MAXN 200005 using namespace std; long tree[MAXN<<2];long a[50005],b[50005];long f[50000][2];long query(long k,long l,long r,long i){ if(k<l || k>r) return 0; if(l<=k && k<=r && tree[i]) return tree[i]; if(l==k && r==k) return tree[i]; long mid = (l+r)>>1; return query(k,l,mid,i<<1)+query(k,mid+1,r,i<<1|1);}void pushdown(long i){ tree[i<<1] = tree[i<<1|1] = tree[i]; tree[i] = 0; return;}void insert(long k,long a,long b,long l,long r,long i){ if(a>r || b<l) return; if(a<=l && r<=b) { tree[i] = k; return; } long mid = (l+r)>>1; if(tree[i]) pushdown(i); insert(k,a,b,l,mid,i<<1); insert(k,a,b,mid+1,r,i<<1|1); return;}int main(){ long n,s,i,j;// FILE* fin;// fin = fopen("d.in","r");// fscanf(fin,"%ld%ld",&n,&s); scanf("%ld%ld",&n,&s); a[0] = b[0] = 0; for(i=n;i>=1;i--) { scanf("%ld%ld",&a[i],&b[i]); // fscanf(fin,"%ld%ld",&a[i],&b[i]); j = query(a[i],-100000,100000,1); f[i][0] = min(abs(a[i]-a[j])+f[j][0],abs(a[i]-b[j])+f[j][1]); j = query(b[i],-100000,100000,1); f[i][1] = min(abs(b[i]-a[j])+f[j][0],abs(b[i]-b[j])+f[j][1]); insert(i,a[i],b[i],-100000,100000,1); } long ans; ans = min(f[1][0]+abs(a[1]-s),f[1][1]+abs(b[1]-s)); printf("%ld\n",ans); // fclose(fin); return 0;}
