Codeforces 527C Glass Carving (最长连续0变形+线段树)

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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:

 

题意是给定一个矩形,不停地纵向或横向切割,问每次切割后,最大的矩形面积是多少。
最大矩形面积=最长的长*最宽的宽
这题,长宽都是10^5,所以,用01序列表示每个点是否被切割,然后,
最长的长就是长的最长连续0的数量+1
最长的宽就是宽的最长连续0的数量+1
于是用线段树维护最长连续零
 
问题转换成:
目标信息:区间最长连续零的个数
点信息:0 或 1
由于目标信息不符合区间加法,所以要扩充目标信息。
 
转换后的线段树结构
区间信息:从左,右开始的最长连续零,本区间是否全零,本区间最长连续零。
点信息:0 或 1
然后还是那2个问题:
 
1.区间加法:
这里,一个区间的最长连续零,需要考虑3部分:
-(1):左子区间最长连续零
-(2):右子区间最长连续零
-(3):左右子区间拼起来,而在中间生成的连续零(可能长于两个子区间的最长连续零)
而中间拼起来的部分长度,其实是左区间从右开始的最长连续零+右区间从左开始的最长连续零。
所以每个节点需要多两个量,来存从左右开始的最长连续零。
然而,左开始的最长连续零分两种情况,
--(1):左区间不是全零,那么等于左区间的左最长连续零
--(2):左区间全零,那么等于左区间0的个数加上右区间的左最长连续零
于是,需要知道左区间是否全零,于是再多加一个变量。
最终,通过维护4个值,达到了维护区间最长连续零的效果。
 
2.点信息->区间信息 : 
如果是0,那么  最长连续零=左最长连续零=右最长连续零=1 ,全零=true。
如果是1,那么  最长连续零=左最长连续零=右最长连续零=0, 全零=false。
 
至于修改和查询,有了区间加法之后,机械地写一下就好了。
由于这里其实只有对整个区间的查询,所以查询函数是不用写的,直接找根的统计信息就行了。

 

递归

技术分享图片
#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
", (query(1, W - 1, 1, 1, W - 1, 0) + 1) * (query(1, H - 1, 1, 1, H - 1, 1) + 1));
        }
    }
}
View Code

非递归

技术分享图片
#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
",(long long)(Max[1][0]+1)*(Max[1][1]+1));
        }
    }
return 0;
}
 
View Code

 

其他解法

https://blog.csdn.net/zearot/article/details/44759437

 





















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