Zoj 1610 Count the Colors (segment tree + interval update + brute force count)

Main topic:

There are n operations, each time is a line of a period of dyeing (color is not the same), sometimes the back color may be covered in front of the color, ask the last color, can see the color there are several, each color has several parts?

Problem Solving Ideas:

This topic makes some difference, not by the point as the object, but with the interval as the object, it is clear that the segment of the tree line operation, update the time to take the interval as a unit, and the calculation of each interval appears several times when the tree can be traversed according to the characteristics of the tree.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5 using namespacestd;6 Const intMAXN =8005;7 structnode8 {9     intL, R, C;Ten     intMid () One     { A         return(R + L)/2; -     } - }; theNode tree[maxn*4]; - intCNT[MAXN], temp; -  - voidBuild (intRootintLintR) + { -TREE[ROOT].L =l; +TREE[ROOT].R =R; ATREE[ROOT].C =-1; at     if(L = = R-1) -         return ; -Build (2*root+1, L, Tree[root]. Mid ()); -Build (2*root+2, Tree[root]. Mid (), r); - } - voidUpdate (intRootintLintRintc) in{//interval Update, -     if(L==TREE[ROOT].L && tree[root].r==R) to     { +TREE[ROOT].C =C; -         return ; the     } *     if(Tree[root].c! =-1) $     {Panax Notoginsengtree[2*root+1].C = tree[2*root+2].C =tree[root].c; -TREE[ROOT].C =-1; the     } +     if(L >=Tree[root]. Mid ()) AUpdate (2*root+2, L, R, c); the     Else if(R <=Tree[root]. Mid ()) +Update (2*root+1, L, R, c); -     Else $     { $Update (2*root+1, L, Tree[root]. Mid (), c); -Update (2*root+2, Tree[root]. Mid (), R, c); -     } the } - voidCount (introot)Wuyi { the     if(Tree[root].c! =-1) -{//The current interval is exposed to the outer surface Wu         if(tree[root].c! = temp)//The current interval has not been counted. -CNT[TREE[ROOT].C] + +; Abouttemp =tree[root].c; $         return ; -     } -     if(TREE[ROOT].L = = tree[root].r-1) -{//returns an area that has not been overwritten, otherwise the stack overflows Atemp =-1; +         return ; the     } -Count (2*root+1); $Count (2*root+2); the } the  the intMain () the { -     intN, M; in      while(SCANF ("%d", &n)! =EOF) the     { theBuild (0,0, MAXN); About  the          while(N--) the         { the             intx, Y, C; +scanf (" %d%d%d", &x, &y, &c); -Update (0, X, Y, c); the         }Bayitemp =-1; thememset (CNT,0,sizeof(CNT)); theCount (0); -          for(intI=0; i<maxn; i++) -             if(Cnt[i]) theprintf ("%d%d\n", I, Cnt[i]); theprintf ("\ n"); the     } the     return 0; -}

Zoj 1610 Count the Colors (segment tree + interval update + brute force count)