Codeforces 527C Glass Carving （最长连续0变形+线段树）
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular w mm × 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?
Input
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.
Output
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Examples
Input
Copy
4 3 4
H 2
V 2
V 3
V 1
Output
Copy
8
4
4
2
Input
Copy
7 6 5
H 4
V 3
V 5
H 2
V 1
Output
Copy
28
16
12
6
4
Note
Picture for the first sample test:
递归
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; const int maxn = 2 * 1e5 + 10; struct SegTree { ll ls, rs, max0; bool is_all0; }segTree[2][maxn<<2]; void pushup(int root, int flag) { SegTree &cur = segTree[flag][root], &lc = segTree[flag][root<<1], &rc = segTree[flag][root<<1|1]; cur.ls = lc.ls + (lc.is_all0 ? rc.ls : 0); cur.rs = rc.rs + (rc.is_all0 ? lc.rs : 0); cur.max0 = max(lc.rs + rc.ls, max(lc.max0, rc.max0)); cur.is_all0 = lc.is_all0 && rc.is_all0; } void build(int L, int R, int root, int flag) { if (L == R) { segTree[flag][root].ls = segTree[flag][root].rs = segTree[flag][root].max0 = 1; segTree[flag][root].is_all0 = true; return; } int mid = (L + R)>>1; build(L, mid, root<<1, flag); build(mid + 1, R, root<<1|1, flag); pushup(root, flag); } void update_node(int L, int R, int root, int pos, int flag) { if (L == R) { segTree[flag][root].ls = segTree[flag][root].rs = segTree[flag][root].max0 = 0; segTree[flag][root].is_all0 = false; return; } int mid = (L + R)>>1; if (pos <= mid) { update_node(L, mid, root<<1, pos, flag); } else { update_node(mid + 1, R, root<<1|1, pos, flag); } pushup(root, flag); } ll query(int L, int R, int root, int qL, int qR, int flag) { if (qL <= L && R <= qR) { return segTree[flag][root].max0; } int mid = (L + R)>>1; ll temp = 0; if (qL <= mid) { temp = max(temp, query(L, mid, root<<1, qL, qR, flag)); } if (qR > mid) { temp = max(temp, query(mid + 1, R, root<<1|1, qL, qR, flag)); } return temp; } int main() { int W, H, q, x; char c[5]; while (scanf("%d %d %d", &W, &H, &q) == 3) { build(1, W - 1, 1, 0); build(1, H - 1, 1, 1); while (q--) { scanf("%s %d", c, &x); if (c[0] == 'V') { update_node(1, W - 1, 1, x, 0); } else { update_node(1, H - 1, 1, x, 1); } printf("%I64d\n", (query(1, W - 1, 1, 1, W - 1, 0) + 1) * (query(1, H - 1, 1, 1, H - 1, 1) + 1)); } } }
非递归
#include <iostream> #include <cstdio> #include <cmath> #define maxn 200001 using namespace std; int L[maxn<<2][2];//从左开始连续零个数 int R[maxn<<2][2];//从右 int Max[maxn<<2][2];//区间最大连续零 bool Pure[maxn<<2][2];//是否全零 int M[2]; void PushUp(int rt,int k){//更新rt节点的四个数据 Pure[rt][k]=Pure[rt<<1][k]&&Pure[rt<<1|1][k]; Max[rt][k]=max(R[rt<<1][k]+L[rt<<1|1][k],max(Max[rt<<1][k],Max[rt<<1|1][k])); L[rt][k]=Pure[rt<<1][k]?L[rt<<1][k]+L[rt<<1|1][k]:L[rt<<1][k]; R[rt][k]=Pure[rt<<1|1][k]?R[rt<<1|1][k]+R[rt<<1][k]:R[rt<<1|1][k]; } void Build(int n,int k){//建树，赋初值 for(int i=0;i<M[k];++i) L[M[k]+i][k]=R[M[k]+i][k]=Max[M[k]+i][k]=Pure[M[k]+i][k]=i<n; for(int i=M[k]-1;i>0;--i) PushUp(i,k); } void Change(int X,int k){//切割，更新 int s=M[k]+X-1; Pure[s][k]=Max[s][k]=R[s][k]=L[s][k]=0; for(s>>=1;s;s>>=1) PushUp(s,k); } int main(void) { int w,h,n; while(cin>>w>>h>>n){ //以下3行，找出非递归线段树的第一个数的位置。 M[0]=M[1]=1; while(M[0]<h-1) M[0]<<=1; while(M[1]<w-1) M[1]<<=1; //建树 Build(h-1,0);Build(w-1,1); for(int i=0;i<n;++i){ //读取数据 char x;int v; scanf(" %c%d",&x,&v); //切割 x=='H'?Change(v,0):Change(v,1); //输出 printf("%I64d\n",(long long)(Max[1][0]+1)*(Max[1][1]+1)); } } return 0; }
其他解法
https://blog.csdn.net/zearot/article/details/44759437
