F-count the Colors-zoj 1610 (interval coverage)

There is a long canvas, now want to draw some color on this canvas, but the color of the painting will cover the color of the front, now want to know the color distribution of the canvas after the painting, such as the number 1th color has several pieces, 2nd color has a few pieces .... Analysis: Basically the poster is the same, but the final request for the output is the color of the painting has a few pieces, Can follow the way of the poster to do it first, and then to each point of inquiry, do not know whether the complexity is too high. But let's try it first. Note: If you see

Segmentation Fault

You can see if the tree is small, and you can't build it.#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
usingnamespaceStd
#defineLson Root<<1,l,tree[root]. Mid ()
#defineRson Root<<1|1,tree[root]. Mid () +1,r

ConstintMAXN =80005;

structhook{intL, R, C;} P[MAXN];
structtree{
intL, R;
intColor//Record Color
intMid () {return(L+R)/2;}
}tree[maxn*4];
intCOLOR[MAXN];

voidCovercolor (intLintRintE//Paint the interval LR color e
{
for(intI=l; i<=r; i++)
Color[i] = e;
}

voidBuild (intRootintLintR
{
Tree[root]. L = L, Tree[root]. R = r;
//0 is not covered by color, 1 is not covered, and 2 means the subtree is covered by another color
Tree[root].color =0;

if(L = = R)return;

Build (Lson);
Build (Rson);
}
voidUp (intRoot
{
if(Tree[root]. L! = Tree[root]. R
if(tree[root<<1].color = =1&& tree[root<<1|1].color = =1)
Tree[root].color =1;
}
voidInsert (intRootintLintRintE
{
if(Tree[root].color = =1)return;

if(Tree[root]. L = = L && tree[root]. r = = R &&!tree[root].color)
{
Tree[root].color =1;
Covercolor (L, R, E);
return;
}
Tree[root].color =2;

if(R <= Tree[root]. Mid ())
Insert (root<<1, L, R, E);
Elseif(L > Tree[root]. Mid ())
Insert (root<<1|1, L, R, E);
Else
{
Insert (Lson, E);
Insert (Rson, E);
}

Up (root);
}
intMain ()
{
intN

while(SCANF ("%d", &n)! = EOF)
{
intI, ans[maxn] = {0};

for(i=1; i<=n; i++)
scanf"%d%d%d", &AMP;P[I].L, &AMP;P[I].R, &p[i].c);

Build (1,0, maxn-1);
memset (Color,-1,sizeof(Color));

for(I=n; i>0; i--)//because a tight interval is given, and the tree points are built, so the left +1 is turned into a dot covering the
Insert (1, p[i].l+1, P[I].R, P[I].C);

for(i=0; i<maxn; i++)
{
if(color[i]!=-1&& (!i | | color[i]!=color[i-1]) )
ans[color[i]]++;
}

for(i=0; i<maxn; i++)if(Ans[i])
printf"%d%d\n", I, ans[i]);
printf"\ n");
}

return0;

} 

F-count the Colors-zoj 1610 (interval coverage)