HDU 3911 Black and White

Black and Whitetime limit:3000msmemory limit:32768kbthis problem'll be judged onHDU. Original id:3911
64-bit integer IO format: %i64d Java class name: Main There is a bunch of stones on the beach; Stone color is white or black. Little Sheep have a magic brush, she can change the color of a continuous stone, black to white, white to black. Little Sheep like black very much, so she want to know the longest period of consecutive black stones in a range [I, j].InputThere was multiple cases, the first line of each case was an integer n (1<= n <= 10^5), followed by n integer 1 or 0 (1 Indicates black stone and 0 indicates white stone), then was an integer M (1<=m<=10^5) followed by M operations forma Tted as X i j (x = 0 or 1), x=1 means change the color of stones in range[i,j], and x=0 means ask the longest period of CO Nsecutive black stones in range[i,j]OutputWhen x=0 output a number means the longest length of black stones in range [I,j].Sample Input

41 0 1 050 1 41 2 30 1 41 3 30 4 4

Sample Output

120

SourceMulti-university Training Contest 8-host by HUST problem solving: line tree ah ... Fortunately only one operation, flip ah ... is to give you two operations, 1 a b indicates that a B area inside the 01 is flipped, 0 A B indicates the output of a B continuous between the longest 1 how many

1#include <bits/stdc++.h>2 using namespacestd;3 Const intMAXN =100010;4 structNode {5     intlsum[2],rsum[2],mx[2];6     BOOLlazy;7} tree[maxn<<2];8 voidPushup (intVintk) {9      for(inti =0; I <2; ++i) {TenTree[v].lsum[i] = tree[v<<1].lsum[i]; OneTree[v].rsum[i] = tree[v<<1|1].rsum[i]; A         if(Tree[v].lsum[i] = = k-(k>>1)) -Tree[v].lsum[i] + = tree[v<<1|1].lsum[i]; -         if(Tree[v].rsum[i] = = (k>>1)) theTree[v].rsum[i] + = tree[v<<1].rsum[i]; -Tree[v].mx[i] = max (tree[v<<1].mx[i],tree[v<<1|1].mx[i]); -Tree[v].mx[i] = max (tree[v].mx[i],tree[v<<1].rsum[i]+tree[v<<1|1].lsum[i]); -     } + } -  + voidChangeintv) { ASwap (tree[v].lsum[0],tree[v].lsum[1]); atSwap (tree[v].rsum[0],tree[v].rsum[1]); -Swap (tree[v].mx[0],tree[v].mx[1]); -Tree[v].lazy =!Tree[v].lazy; - } - voidPushdown (intv) { -     if(tree[v].lazy) { inChange (v<<1); -Change (v<<1|1); toTree[v].lazy =false; +     } - } the voidBuildintLintRintv) { *Tree[v].lazy =false; $     if(L = =R) {Panax Notoginseng         inttmp; -scanf"%d",&tmp); thetree[v].mx[0] = tree[v].lsum[0] = tree[v].rsum[0] = !tmp; +tree[v].mx[1] = tree[v].lsum[1] = tree[v].rsum[1] =tmp; A         return; the     } +     intMid = (L + R) >>1; -Build (l,mid,v<<1); $Build (mid+1,r,v<<1|1); $Pushup (V,r-l +1); - } - voidUpdateintLintRintLtintRtintv) { the     if(LT <= L && RT >=R) { - Change (v);Wuyi         return; the     } - Pushdown (v); Wu     intMid = (L + R) >>1; -     if(LT <= mid) Update (l,mid,lt,rt,v<<1); About     if(Rt > Mid) Update (mid+1,r,lt,rt,v<<1|1); $Pushup (V,r-l +1); - } - intQueryintLintRintLtintRtintv) { -     if(LT <= L && RT >= R)returntree[v].mx[1]; A Pushdown (v); +     intMid = (L + R) >>1, ret =0; the     if(LT <= mid) ret = max (Ret,query (l,mid,lt,rt,v<<1)); -     if(Rt > Mid) ret = max (Ret,query (mid+1,r,lt,rt,v<<1|1)); $     if(LT <= mid && rt >mid) theret = max (Ret,min (Mid-lt +1,tree[v<<1].rsum[1]) + min (rt-mid,tree[v<<1|1].lsum[1])); thePushup (V,r-l +1); the     returnret; the } - intMain () { in     intn,m,op,a,b; the      while(~SCANF ("%d",&N)) { theBuild1N1); Aboutscanf"%d",&m); the          while(m--){ thescanf" %d%d%d",&op,&a,&b); the             if(OP) Update (1, N,a,b,1); +             Elseprintf"%d\n", Query (1, N,a,b,1)); -         } the     }Bayi     return 0; the}

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HDU 3911 Black and White