題目鏈接
http://codeforces.com/contest/1202/problem/C
題目描述
You have a string ss — a sequence of commands for your toy robot. The robot is placed in some cell of a rectangular grid. He can perform four commands:
- 'W' — move one cell up;
- 'S' — move one cell down;
- 'A' — move one cell left;
- 'D' — move one cell right.
Let Grid(s) be the grid of minimum possible area such that there is a position in the grid where you can place the robot in such a way that it will not fall from the grid while running the sequence of commands. For example, if s=DSAWWAW then Grid(s) is the 4×3 grid:
- you can place the robot in the cell (3,2);
- the robot performs the command 'D' and moves to (3,3);
- the robot performs the command 'S' and moves to (4,3);
- the robot performs the command 'A' and moves to (4,2);
- the robot performs the command 'W' and moves to (3,2);
- the robot performs the command 'W' and moves to (2,2);
- the robot performs the command 'A' and moves to (2,1);
- the robot performs the command 'W' and moves to (1,1).
You have 44 extra letters: one 'W', one 'A', one 'S', one 'D'. You'd like to insert at most one of these letters in any position of sequence ss to minimize the area of Grid(s).
What is the minimum area of Grid(s) you can achieve?
Input
The first line contains one integer TT (1≤T≤1000) — the number of queries.
Next TT lines contain queries: one per line. This line contains single string ss (1≤|s|≤2⋅10^5, si∈{W,A,S,D}) — the sequence of commands.
It's guaranteed that the total length of ss over all queries doesn't exceed 2⋅10^5
Output
Print T integers: one per query. For each query print the minimum area of Grid(s) you can achieve.
Example
input
3
DSAWWAW
D
WA
output
8
2
4
題意
給一個字符串序列包含{W,S,A,D}四個字符,分別表示在某一個初始位置向上下左右移動一個格子。
定義Grid(s)爲 i一個包含所有所走位置的最小面積矩形。現在在該字符串上最多插入一個字符，問能構成的最小Grid(s)
思路
將當前位置的橫縱座標分開，分別用兩個線段樹維護各自的最大值與最小值。如要插入字符則將區間(i+1,n)所有元素加1或減1。
代碼雖然很長但大部分都是直接套的模板。
代碼
#include<bits/stdc++.h>
#define ll long long
#define maxn 200010
using namespace std;
ll n;
const ll MAXM=200010;
ll a[MAXM+5],smax[(MAXM<<2)+5],smin[(MAXM<<2)+5],add[(MAXM<<2)+5];
ll b[MAXM+5],smax1[(MAXM<<2)+5],smin1[(MAXM<<2)+5],add1[(MAXM<<2)+5];
void init(){
ll i,len=n*4+5;
for(i=0;i<len;i++){
smax[i]=smin[i]=add[i]=smax1[i]=smin1[i]=add1[i]=0;
}
for(i=0;i<n+3;i++)a[i]=b[i]=0;
}
void build(ll o,ll l,ll r){
if(l==r)smin[o]=smax[o]=a[l];
else{
ll m=l+((r-l)>>1);
build(o<<1,l,m);
build((o<<1)|1,m+1,r);
smax[o]=max(smax[o<<1],smax[(o<<1)|1]);
smin[o]=min(smin[o<<1],smin[(o<<1)|1]);
}
}
void pushup(ll o){
smin[o]=min(smin[o<<1],smin[(o<<1)|1]);
smax[o]=max(smax[o<<1],smax[(o<<1)|1]);
}
void pushdown(ll o,ll l,ll r){
if(add[o]){
add[o<<1]+=add[o];
add[o<<1|1]+=add[o];
smin[o<<1]+=add[o];
smin[o<<1|1]+=add[o];
smax[o<<1]+=add[o];
smax[o<<1|1]+=add[o];
add[o]=0;
}
}
void update(ll o,ll l,ll r,ll ql,ll qr,ll addv){
if(ql<=l&&qr>=r){
add[o]+=addv;
smin[o]+=addv;
smax[o]+=addv;
return;
}
pushdown(o,l,r);
ll m=l+((r-l)>>1);
if(ql<=m)update(o<<1,l,m,ql,qr,addv);
if(qr>=m+1)update(o<<1|1,m+1,r,ql,qr,addv);
pushup(o);
}
ll query_max(ll o,ll l,ll r,ll ql,ll qr){
if(ql<=l&&qr>=r) return smax[o];
pushdown(o,l,r);
ll m=l+((r-l)>>1);
ll ans=0;
if(ql<=m)ans=query_max(o<<1,l,m,ql,qr);
if(qr>=m+1)ans=max(ans,query_max(o<<1|1,m+1,r,ql,qr));
return ans;
}
ll query_min(ll o,ll l,ll r,ll ql,ll qr){
if(ql<=l&&qr>=r) return smin[o];
pushdown(o,l,r);
ll m=l+((r-l)>>1);
ll ans=0;
if(ql<=m)ans=query_min(o<<1,l,m,ql,qr);
if(qr>=m+1)ans=min(ans,query_min(o<<1|1,m+1,r,ql,qr));
return ans;
}
void build1(ll o,ll l,ll r){
if(l==r)smin1[o]=smax1[o]=b[l];
else{
ll m=l+((r-l)>>1);
build1(o<<1,l,m);
build1((o<<1)|1,m+1,r);
smax1[o]=max(smax1[o<<1],smax1[(o<<1)|1]);
smin1[o]=min(smin1[o<<1],smin1[(o<<1)|1]);
}
}
void pushup1(ll o){
smin1[o]=min(smin1[o<<1],smin1[(o<<1)|1]);
smax1[o]=max(smax1[o<<1],smax1[(o<<1)|1]);
}
void pushdown1(ll o,ll l,ll r){
if(add1[o]){
add1[o<<1]+=add1[o];
add1[o<<1|1]+=add1[o];
smin1[o<<1]+=add1[o];
smin1[o<<1|1]+=add1[o];
smax1[o<<1]+=add1[o];
smax1[o<<1|1]+=add1[o];
add1[o]=0;
}
}
void update1(ll o,ll l,ll r,ll ql,ll qr,ll add1v){
if(ql<=l&&qr>=r){
add1[o]+=add1v;
smin1[o]+=add1v;
smax1[o]+=add1v;
return;
}
pushdown1(o,l,r);
ll m=l+((r-l)>>1);
if(ql<=m)update1(o<<1,l,m,ql,qr,add1v);
if(qr>=m+1)update1(o<<1|1,m+1,r,ql,qr,add1v);
pushup1(o);
}
ll query_max1(ll o,ll l,ll r,ll ql,ll qr){
if(ql<=l&&qr>=r) return smax1[o];
pushdown1(o,l,r);
ll m=l+((r-l)>>1);
ll ans=0;
if(ql<=m)ans=query_max1(o<<1,l,m,ql,qr);
if(qr>=m+1)ans=max(ans,query_max1(o<<1|1,m+1,r,ql,qr));
return ans;
}
ll query_min1(ll o,ll l,ll r,ll ql,ll qr){
if(ql<=l&&qr>=r) return smin1[o];
pushdown1(o,l,r);
ll m=l+((r-l)>>1);
ll ans=0;
if(ql<=m)ans=query_min1(o<<1,l,m,ql,qr);
if(qr>=m+1)ans=min(ans,query_min1(o<<1|1,m+1,r,ql,qr));
return ans;
}
char s[maxn];
int main(){
ll t,i;
cin>>t;
memset(smax,0,sizeof(smax));
memset(smin,0,sizeof(smin));
memset(add,0,sizeof(add));
memset(smax1,0,sizeof(smax1));
memset(smin1,0,sizeof(smin1));
memset(add1,0,sizeof(add1));
while(t--){
scanf("%s",s);
n=strlen(s)+1;
a[1]=b[1]=0;
for(i=0;i<strlen(s);i++){
if(s[i]=='A'){
a[i+2]=a[i+1]+1;
b[i+2]=b[i+1];
}
if(s[i]=='D'){
a[i+2]=a[i+1]-1;
b[i+2]=b[i+1];
}
if(s[i]=='W'){
a[i+2]=a[i+1];
b[i+2]=b[i+1]+1;
}
if(s[i]=='S'){
a[i+2]=a[i+1];
b[i+2]=b[i+1]-1;
}
}
build(1,1,n);
build1(1,1,n);
ll al,ar,bl,br,ans;
ll sal,sar,sbl,sbr;
sal=query_min(1,1,n,1,n);
sar=query_max(1,1,n,1,n);
sbl=query_min1(1,1,n,1,n);
sbr=query_max1(1,1,n,1,n);
ans=(sar-sal+1)*(sbr-sbl+1);
for(i=1;i<=n-1;i++){
update(1,1,n,i+1,n,1);
al=query_min(1,1,n,1,n);
ar=query_max(1,1,n,1,n);
if(a[i]+1>ar)ar=a[i]+1;
bl=sbl;
br=sbr;
ans=min(ans,(ar-al+1)*(br-bl+1));
update(1,1,n,i+1,n,-2);
al=query_min(1,1,n,1,n);
ar=query_max(1,1,n,1,n);
if(a[i]-1<al)al=a[i]-1;
ans=min(ans,(ar-al+1)*(br-bl+1));
update(1,1,n,i+1,n,1);
update1(1,1,n,i+1,n,1);
al=sal;
ar=sar;
bl=query_min1(1,1,n,1,n);
br=query_max1(1,1,n,1,n);
if(b[i]+1>br)br=b[i]+1;
ans=min(ans,(ar-al+1)*(br-bl+1));
update1(1,1,n,i+1,n,-2);
bl=query_min1(1,1,n,1,n);
br=query_max1(1,1,n,1,n);
if(b[i]-1<bl)bl=b[i]-1;
ans=min(ans,(ar-al+1)*(br-bl+1));
update1(1,1,n,i+1,n,1);
}
printf("%lld\n",ans);
init();
}
return 0;
}