blue tree monitor

The answer sequence must be a lower convex hull, so the addition of the arithmetic progression is a single-peak function, you can first calculate the maximum value, then two to find out 0 points, and then use the segment tree, the resulting range is modified to a arithmetic progression.This should be better to think about it, although relatively slow ...#include 　　bzoj1568: [Jsoi2008]blue Mary opens company

Previous questions children queue time limit: 1.0s memory limit: 256.0MBProblem description N A child stands in a row. Now it's time to put them in the order of height from low to high, but only two children are exchanged at each location.Every child has a degree of upset. At the beginning, all the children's unhappy degree is 0.If a child is asked to exchange for the first time, his degree of upset increases by 1, and if he is asked to exchange for the second time, his degree of upset increases

of input contains two integers n, m, respectively, the number of important locations in C city and the number of road lines that can be constructed. All locations are numbered from 1 to n sequentially.The next M-line, three integers a, B, c for each line, indicates that a road from location A to location B can be built, costing C. If c is positive, it means that the construction is spending money, if C is negative, then you can make money after building the road (such as building toll roads).Th

Resource contract: Peak memory consumption #include"Cstdio"#include"Cmath"#include"algorithm"using namespacestd;Const intmaxn=100005; typedefLong LongLL;structnode{intL,r; LL sum;} SEGTREE[MAXN*3];intCNT;voidBuildintRtintLintR) {SEGTREE[RT].L=l; SEGTREE[RT].R=R; if(l==r) {scanf ("%lld",segtree[rt].sum); if(segtree[rt].sum==1) {CNT++;//count the number of 1 to facilitate optimization of the programsegtree[rt].sum++;//Change all 1 to 2 to prevent 1 interference program Optimization }

return 0; the } - Else if(Xy) { Wu returnSum[i]; - } About intMid= (L+R)/2; $ intlsum=0, rsum=0; - if(mid>=x) { -Lsum=querysum (i*2, l,mid,x,y); - } A if(midy) { +Rsum=querysum (i*2+1, mid+1, r,x,y); the } - returnlsum+rsum; $ } the the intQueryma (intIintLintRintXinty) the { the if(X>r | | yl) { - return 0; in } the Else if(Xy) { the returnMa[i]; About } the intMid= (L+R)/2; the intLma=0, rma=0; th

Photoshop adds a dark blue backlight to the tree-side characters Material picture color palette, direct the dark and corner area of the dark, and then use curve to increase the dark part of the blue, and then in the character head area rendering a little warm, high light, simulate the backlight effect can be. Original Final effect

much or not clear, and then on the Internet to learn from the great God code, just understand how.1#include 2#include 3#include 4 using namespacestd;5 structnode{6 ints,e,v;7}edge[100010];8 intm=0, n,val[10010],p;9 voidAddintSintEintv) {Tenedge[++m].s=s; OneEdge[m].e=e; Aedge[m].v=v; - } - intfather[10010]; the intFINDF (inti) {//and check set, intends to use Kruskal to calculate the minimum spanning tree; - if(father[i]==i)returni; - Els

There is a data of height of 0 ...The principle of reverse order in a tree-like array:Add (h[j],1); The number of occurrences of the data with height of h[j] plus 1SUM (h[j]);//IWhich is the inverse logarithm#include"Cstdio"#include"CString"#defineLowbit (i) i (-i)using namespacestd;Const intmaxn=1000105;//determined by the range of HtypedefLong LongLL;intBIT[MAXN];voidAddintIintx) { while(iMAXN) {Bit[i]+=x; I+=lowbit (i); }}intSuminti) { intan

Test instructionsN Cities, M-bars. In addition, each city can also build a pier.If there are docks in two cities, these two cities can reach each other.To find the minimum spanning tree.Ideas:For the pier, you can connect all the terminals to a 0 point, at the cost of building a pier.At this time run two times, again there is a pier situation, the second time there is no pier.In this case, the question is whether the road will be connected without a pier.Then also note that negative roads must b