CodeForce 527 C.Glass Carving線分樹

7612 ワード
線分樹
http://codeforces.com/contest/527/problem/C
 
C.Glass Carving
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
out put
スタンダードアウト
Leonid wants to become a glass carver.He already has a rectanglar w mm×h mm sheet of glass,a diamond glass cutter and lots of enthussiasm.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 ntire sheet.This process recult in making smaking smarectlant the frand.Legladmedeglant.frand.a cut divides each fragment of glass that it goes through into smaler frame nts.
After each cut Leonid tries to determine what are the largest of the currently available glass frass frants has.Since there appar more and more framents,this question Tared more and more and more and dime and direme and distract the procements frame frame fretment frame.frame.
Leonid offrs to divide the labor—he will cut glass、and you will calculate the ara of the maximframe after each cut.Dougyou agree?
Input
The first line contains three integers w、?h、?n(2̵≦?w、?h≦?200?、1?n≦?n≦
Next n LINE 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 the distancy milimeters(1?≦??hh h h-?1)fm the lower eddge of theorororororininininininininininininininshshshshshshshshshes s s s s s s s s s s.Inconconconconconconconvers s s s((((8200 1 820 1 820 1)820 0 0 a a a a a a a a a a a a a a a a a a a a a a a a a 820 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0≦̵w̵-̵1)milimeters from the left edge of the original sheet of glass.It is garanted that Leonide won't make two identical cuts.
Output
After each cut print on a single line the ara of the maximbable glass frame in mm 2.
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
ノート
Picture for the first sample test:

Picture for the second sample test:

 
 
線分樹は、切断されていない行を0で表し、すでに切断されたものを1で表します.行列の最長連続0個の数を求めます.
 
ACコード
#include 
#include
using namespace std;

#define maxn 200007
struct node1{
    long long left=0;
    long long right=0;
    long long cnt=0;
    long long flag=0;
};
struct node2{
    long long left=0;
    long long right=0;
    long long cnt=0;
    long long flag=0;
};
node1 nodes1[maxn<<2];
node2 nodes2[maxn<<2];

void pushup1(int rt)
{
    if(nodes1[rt<<1].flag==0&&nodes1[rt<<1|1].flag==0)
    {
        nodes1[rt].left=nodes1[rt<<1].left+nodes1[rt<<1|1].left;
        nodes1[rt].right=nodes1[rt].left;
        nodes1[rt].cnt=0;
        nodes1[rt].flag=0;
        return;
    }
    if(nodes1[rt<<1].flag==0&&nodes1[rt<<1|1].flag==1)
    {
        nodes1[rt].left=nodes1[rt<<1].left+nodes1[rt<<1|1].left;
        nodes1[rt].right=nodes1[rt<<1|1].right;
        nodes1[rt].cnt=nodes1[rt<<1|1].cnt;
        nodes1[rt].flag=1;
    }
    if(nodes1[rt<<1].flag==1&&nodes1[rt<<1|1].flag==0)
    {
        nodes1[rt].right=nodes1[rt<<1|1].right+nodes1[rt<<1].right;
        nodes1[rt].left=nodes1[rt<<1].left;
        nodes1[rt].cnt=nodes1[rt<<1].cnt;
        nodes1[rt].flag=1;
    }
    if(nodes1[rt<<1].flag==1&&nodes1[rt<<1|1].flag==1)
    {
        nodes1[rt].left=nodes1[rt<<1].left;
        nodes1[rt].right=nodes1[rt<<1|1].right;
        nodes1[rt].cnt=max(nodes1[rt<<1].right+nodes1[rt<<1|1].left,max(nodes1[rt<<1].cnt,nodes1[rt<<1|1].cnt));
        nodes1[rt].flag=1;
    }
}

void pushup2(int rt)
{
    if(nodes2[rt<<1].flag==0&&nodes2[rt<<1|1].flag==0)
    {
        nodes2[rt].left=nodes2[rt<<1].left+nodes2[rt<<1|1].left;
        nodes2[rt].right=nodes2[rt].left;
        nodes2[rt].cnt=0;
        nodes2[rt].flag=0;
        return;
    }
    if(nodes2[rt<<1].flag==0&&nodes2[rt<<1|1].flag==1)
    {
        nodes2[rt].left=nodes2[rt<<1].left+nodes2[rt<<1|1].left;
        nodes2[rt].right=nodes2[rt<<1|1].right;
        nodes2[rt].cnt=nodes2[rt<<1|1].cnt;
        nodes2[rt].flag=1;
    }
    if(nodes2[rt<<1].flag==1&&nodes2[rt<<1|1].flag==0)
    {
        nodes2[rt].right=nodes2[rt<<1|1].right+nodes2[rt<<1].right;
        nodes2[rt].left=nodes2[rt<<1].left;
        nodes2[rt].cnt=nodes2[rt<<1].cnt;
        nodes2[rt].flag=1;
    }
    if(nodes2[rt<<1].flag==1&&nodes2[rt<<1|1].flag==1)
    {
        nodes2[rt].left=nodes2[rt<<1].left;
        nodes2[rt].right=nodes2[rt<<1|1].right;
        nodes2[rt].cnt=max(nodes2[rt<<1].right+nodes2[rt<<1|1].left,max(nodes2[rt<<1].cnt,nodes2[rt<<1|1].cnt));
        nodes2[rt].flag=1;
    }
}
void build1(int l,int r,int rt)
{
    if(l==r)
    {
        nodes1[rt].left=1;
        nodes1[rt].right=1;
        nodes1[rt].cnt=0;
        nodes1[rt].flag=0;
        return;
    }
    else
    {
        int mid=(l+r)>>1;
        build1(l,mid,rt<<1);
        build1(mid+1,r,rt<<1|1);
        pushup1(rt);
    }
}

void build2(int l,int r,int rt)
{
    if(l==r)
    {
        nodes2[rt].left=1;
        nodes2[rt].right=1;
        nodes2[rt].cnt=0;
        nodes2[rt].flag=0;
        return;
    }
    else
    {
        int mid=(l+r)>>1;
        build2(l,mid,rt<<1);
        build2(mid+1,r,rt<<1|1);
        pushup2(rt);
    }
}

void update1(int L,int C,int l,int r,int rt)
{
    if(l==r)
    {
        nodes1[rt].left=0;
        nodes1[rt].right=0;
        nodes1[rt].cnt=0;
        nodes1[rt].flag=1;
        return;
    }
    else
    {
        int mid=(l+r)>>1;
        if(L<=mid)
            update1(L,C,l,mid,rt<<1);
        else
            update1(L,C,mid+1,r,rt<<1|1);
        pushup1(rt);
    }
}

void update2(int L,int C,int l,int r,int rt)
{
    if(l==r)
    {
        nodes2[rt].left=0;
        nodes2[rt].right=0;
        nodes2[rt].cnt=0;
        nodes2[rt].flag=1;
        return;
    }
    else
    {
        int mid=(l+r)>>1;
        if(L<=mid)
            update2(L,C,l,mid,rt<<1);
        else
            update2(L,C,mid+1,r,rt<<1|1);
        pushup2(rt);
    }
}

int main()
{
    std::ios::sync_with_stdio(false);
    long long w,h,n;
    cin>>w>>h>>n;
    build1(1,w,1);
    build2(1,h,1);
    for(int i=1;i<=n;++i)
    {
        char a;
        int b;
        cin>>a>>b;
        if(a=='V')
        {
            update1(b,1,1,w,1);
            long long temp1=max(nodes1[1].left+1,max(nodes1[1].cnt+1,nodes1[1].right));

            long long temp2=max(nodes2[1].cnt+1,max(nodes2[1].left+1,nodes2[1].right));
            cout<
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