Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular wmm × h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
The first line contains three integers w, h, n (2 ≤ w, h ≤ 200 000, 1 ≤ n ≤ 200 000).
Next n lines contain the descriptions of the cuts. Each description has the form H y or V x. In the first case Leonid makes the horizontal cut at the distance y millimeters (1 ≤ y ≤ h - 1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance x (1 ≤ x ≤ w - 1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
After each cut print on a single line the area of the maximum available glass fragment in mm2.
4 3 4
H 2
V 2
V 3
V 1
8
4
4
2
7 6 5
H 4
V 3
V 5
H 2
V 1
28
16
12
6
4
Picture for the first sample test:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 #include <cmath> 8 #include <set> 9 using namespace std; 10 11 #define N 200005 12 #define ll root<<1 13 #define rr root<<1|1 14 #define mid1 (a[root].l+a[root].r)/2 15 #define mid2 (b[root].l+b[root].r)/2 16 17 int w, h, n; 18 struct node{ 19 int l, r; 20 int lx, rx; 21 int maxh; 22 }a[N*4], b[N*4]; 23 24 void build1(int l,int r,int root){ 25 a[root].l=l; 26 a[root].r=r; 27 if(l==r){ 28 if(l==0) a[root].lx=a[root].rx=0; 29 else if(l==w) a[root].lx=a[root].rx=w; 30 else a[root].lx=a[root].rx=-1; 31 a[root].maxh=0; 32 return; 33 } 34 build1(l,mid1,ll); 35 build1(mid1+1,r,rr); 36 a[root].lx=a[ll].lx; 37 a[root].rx=a[rr].rx; 38 a[root].maxh=a[root].r-a[root].l; 39 } 40 41 void build2(int l,int r,int root){ 42 b[root].l=l; 43 b[root].r=r; 44 if(l==r){ 45 if(l==0) b[root].lx=b[root].rx=0; 46 else if(l==h) b[root].lx=b[root].rx=h; 47 else b[root].lx=b[root].rx=-1; 48 b[root].maxh=0; 49 return; 50 } 51 build2(l,mid2,ll); 52 build2(mid2+1,r,rr); 53 b[root].lx=b[ll].lx; 54 b[root].rx=b[rr].rx; 55 b[root].maxh=b[root].r-b[root].l; 56 } 57 58 void update1(int p,int root){ 59 if(a[root].l==p&&a[root].r==p){ 60 a[root].lx=a[root].rx=p;return; 61 } 62 if(p<=a[ll].r) update1(p,ll); 63 else update1(p,rr); 64 int ln, rn; 65 if(a[ll].lx!=-1) a[root].lx=a[ll].lx; 66 else if(a[ll].rx!=-1) a[root].lx=a[ll].rx; 67 else if(a[rr].lx!=-1) a[root].lx=a[rr].lx; 68 else if(a[rr].rx!=-1) a[root].lx=a[rr].rx; 69 else a[root].lx=-1; 70 71 if(a[rr].rx!=-1) a[root].rx=a[rr].rx; 72 else if(a[rr].lx!=-1) a[root].rx=a[rr].lx; 73 else if(a[ll].rx!=-1) a[root].rx=a[ll].rx; 74 else if(a[ll].lx!=-1) a[root].rx=a[ll].lx; 75 else a[root].rx=-1; 76 77 if(a[ll].rx!=-1) ln=a[ll].r-a[ll].rx; 78 else ln=a[ll].r-a[ll].l; 79 if(a[rr].lx!=-1) rn=a[rr].lx-a[rr].l; 80 else rn=a[rr].r-a[rr].l; 81 82 a[root].maxh=max(max(a[ll].maxh,a[rr].maxh),ln+rn+1); 83 } 84 85 void update2(int p,int root){ 86 if(b[root].l==p&&b[root].r==p){ 87 b[root].lx=b[root].rx=p;return; 88 } 89 if(p<=b[ll].r) update2(p,ll); 90 else update2(p,rr); 91 int ln, rn; 92 if(b[ll].lx!=-1) b[root].lx=b[ll].lx; 93 else if(b[ll].rx!=-1) b[root].lx=b[ll].rx; 94 else if(b[rr].lx!=-1) b[root].lx=b[rr].lx; 95 else if(b[rr].rx!=-1) b[root].lx=b[rr].rx; 96 else b[root].lx=-1; 97 98 if(b[rr].rx!=-1) b[root].rx=b[rr].rx; 99 else if(b[rr].lx!=-1) b[root].rx=b[rr].lx; 100 else if(b[ll].rx!=-1) b[root].rx=b[ll].rx; 101 else if(b[ll].lx!=-1) b[root].rx=b[ll].lx; 102 else b[root].rx=-1; 103 if(b[ll].rx!=-1) ln=b[ll].r-b[ll].rx; 104 else ln=b[ll].r-b[ll].l; 105 if(b[rr].lx!=-1) rn=b[rr].lx-b[rr].l; 106 else rn=b[rr].r-b[rr].l; 107 b[root].maxh=max(max(b[ll].maxh,b[rr].maxh),ln+rn+1); 108 } 109 110 main() 111 { 112 int i, j, k; 113 while(scanf("%d %d %d",&w,&h,&n)==3){ 114 char s[5]; 115 build1(0,w,1); 116 build2(0,h,1); 117 // printf("%d %d ",a[1].maxh,b[1].maxh); 118 while(n--){ 119 scanf("%s%d",s,&k); 120 if(s[0]=='H'){ 121 update2(k,1); 122 } 123 else{ 124 update1(k,1); 125 } 126 // printf("%d %d ",a[1].maxh,b[1].maxh); 127 printf("%I64d ",(__int64)a[1].maxh*(__int64)b[1].maxh); 128 } 129 } 130 }