[DTOJ # 1772] transportation task / [] RIJEKA COCI2011OPEN

Idea: Just consider the destinations on the left departure v u, ie u> v of the situation. A total of tot personal note of departure and destination to satisfy u> v.

The request interval [v, u] and set as to minimize the total distance of the U-turn. The answer is $ m + \ cup_ {i = 1} ^ {tot} [v_i, u_i] $.

May seek union by consolidating interval the way, is about to have a cross section into a new range.

note:

bool cmp(node x,node y){return x.l<y.l||(x.l==y.l&&x.r<y.r);}

 

sort(b+1,b+1+t2,cmp);
d[t4=1]=b[1];
for(i=2;i<=t2;++i)
{
	if (jiao(d[t4],b[i]))
	{
		d[t4].r=Max(d[t4].r,b[i].r);
		d[t4].l=Min(d[t4].l,b[i].l);
	}
	else d[++t4]=b[i];
}

By the intersection of the above method, i.e., the interval will merge sort order from front to rear, left to right ordering required. Right-to-left sort may arise following:

| A |    | four |

    | Two |        | three |

All of these can be combined into a range, but when scanning, one and two combined, three and four merge, duplicate statistics intersection of two and four.

If you press the right-to-left sort, you need to merge from back to front. "After the front left and right."