C. Glass Carving
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangularwmm×hmm 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 integersw,h,n(2≤w,h≤200000,1≤n≤200000).
Nextnlines contain the descriptions of the cuts. Each description has the formHyorVx. In the first case Leonid makes the horizontal cut at the distanceymillimeters (1≤y≤h-1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distancex(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.
Sample test(s)
input
4 3 4H 2V 2V 3V 1
output
8442
input
7 6 5H 4V 3V 5H 2V 1
output
28161264
Note
Picture for the first sample test:
Picture for the second sample test:
题意:切玻璃，每切一刀求当前的最大快的面积
思路: 最重要到额一点是找横的最大和竖着的最大
如果用 set,每次把新的长宽插入，，然后erase（）被破坏的长或者宽
如果用并查集，先把没有切的看成一个整体，然后重后往前依次去掉一条切得线后可以得到新的一个长或者宽
multiset 代码：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n)scanf("%s", n)#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 10005multiset<ll>x,y;multiset<ll>hx,hy;multiset<ll>::iterator it;char ch[10];ll w,h;ll xx;int n;int main(){int i,j;scanf("%I64d%I64d%d",&w,&h,&n);x.insert(0);x.insert(w);y.insert(0);y.insert(h);hx.insert(w);hy.insert(h);while(n–){scanf("%s%d",ch,&xx);if(ch[0]=='V'){x.insert(xx);it=x.find(xx);it++;ll ri=*it;it–;it–;ll le=*it;it=hx.find(ri-le); //这一行和下一行不能用hx.erase(ri-le)代替，因为这样是删除所有的hx.erase(it);//插入一条线，破坏了一个长度hx.insert(ri-xx); //增加了两个长度hx.insert(xx-le);}else{y.insert(xx);it=y.find(xx);it++;ll ri=*it;it–;it–;ll le=*it;it=hy.find(ri-le);hy.erase(it);hy.insert(ri-xx);hy.insert(xx-le);}it=hx.end();it–;ll le=*it;it=hy.end();//长宽都取最大值it–;ll ri=*it;pf("%I64d\n",le*ri);}return 0;}
并查集代码：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n)scanf("%s", n)#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 200005int lenh,lenw; //长的最大值与宽的最大值int fah[N],faw[N];int numh[N],numw[N];int vish[N],visw[N];int w,h,n;int op[N];ll ans[N];char ch[N][3];int chah(int x){ if(fah[x]!=x)fah[x]=chah(fah[x]); return fah[x];}void unith(int x,int y){x=chah(x);y=chah(y);if(x==y) return ;fah[y]=x; numh[x]+=numh[y]; lenh=max(numh[x],lenh);}int chaw(int x){ if(faw[x]!=x)faw[x]=chaw(faw[x]); return faw[x];}void unitw(int x,int y){x=chaw(x);y=chaw(y);if(x==y) return ;faw[y]=x; numw[x]+=numw[y];lenw=max(numw[x],lenw);}int main(){int i,j;sfff(w,h,n);mem(visw,0);mem(vish,0);int t=max(h,w);fre(i,0,t+1){fah[i]=i;numh[i]=1;faw[i]=i;numw[i]=1;}numh[0]=numw[0]=0;int x;fre(i,0,n){scanf("%s%d",ch[i],&op[i]);if(ch[i][0]=='H')vish[op[i]]=1;elsevisw[op[i]]=1;}lenh=1;lenw=1;fre(i,1,h)if(!vish[i])unith(i,i+1);fre(i,1,w)if(!visw[i])unitw(i,i+1);// fre(i,1,h)//if(fah[i]==i)// {// pf("%d %d\n",i,numh[i]);//// }// fre(i,1,w)//if(faw[i]==i)// {// pf("%d %d\n",i,numw[i]);//// }free(i,n-1,0){//pf("****%d %d\n",lenw,lenh);ans[i]=(ll)(lenh)*(ll)(lenw);if(ch[i][0]=='H')unith(op[i],op[i]+1);elseunitw(op[i],op[i]+1);}fre(i,0,n){pf("%I64d\n",ans[i]);} return 0;}
回首往事，日子里竟全是斑澜的光影，
