QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#211119 | #5414. Stop, Yesterday Please No More | ucup-team1001 | RE | 0ms | 3876kb | C++14 | 2.6kb | 2023-10-12 10:18:20 | 2023-10-12 10:18:22 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using ll = long long;
#define int ll
#define endl "\n"
#define no cout << "NO" << endl
#define yes cout << "YES" << endl
#define pb push_back
#define mkp make_pair
#define fi first
#define se second
#define ary(i) array<ll,(i)>
#define all(x) x.begin(),x.end()
const int maxn=2e5+7;
const int inf=1e9+7;
const int mod=998244353;
void solve(){
// 高为n,宽为m, 最后需要 剩下
int n,m,k;
cin>>n>>m>>k;
string s;
cin>>s;
// 记录可能的偏移量 移除偏移量后就是活着的
int maxx=0,maxy=0,minx=0,miny=0;
int x=0,y=0;
// 模拟上下左右移动
for(auto i:s){
if(i=='U')y++;
else if(i=='D')y--;
else if(i=='L')x--;
else x++;
// 记录最大的点
maxx=max(x,maxx);
maxy=max(y,maxy);
minx=min(minx,x);
miny=min(miny,y);
}
// 计算可能的上下边界
int D = -miny;
int U = n-1-maxy;
int L = -minx;
int R = m-1- maxx;
//判定全死
if(D>U||L>R){
if(k==0){
cout<<m*n<<endl;
}else {
cout<<0<<endl;
}
return;
}
// 计算死亡的袋鼠数
int die=(U-D+1)*(R-L+1)-k;
// cerr<<die<<endl;
// 开二倍图
vector<vector<int>> v(2*n+5,vector<int>(2*m+5,0));
int primex=n+1;
int primey=m+1;
v[primex][primey]=1;
x=primex;
y=primey;
// 跑图
for(auto i:s){
if(i=='U')y++;
else if(i=='D')y--;
else if(i=='L')x--;
else x++;
// 记录最大的点
v[x][y]=1;
}
// 进行前缀和
for(int i =0;i<2*n+3;i++){
for(int j=1;j<2*m+3;j++){
v[i][j]+=v[i][j-1];
}
}
for(int i=0;i<2*m+3;i++){
for(int j=1;j<2*n+3;j++){
v[j][i]+=v[j-1][i];
}
}
int ans =0;
// cerr<<U<<" "<<D<<" "<<L<<" "<<R<<endl;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
int localx=primex-j;
int localy=primey-i;
int t=v[localy+U][localx+R]-
v[localy+D-1][localx+R]-
v[localy+U][localx+L-1]+
v[localy+D-1][localx+L-1];
// cerr<<t<<" ";
if(t==die){
ans++;
}
}
// cerr<<endl;
}
cout<<ans<<endl;
}
#undef int
int main()
{
IOS
int n;
cin>>n;
// n=1;
while(n--){
solve();
}
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3876kb
input:
3 4 5 3 ULDDRR 4 5 0 UUUUUUU 4 5 10 UUUUUUU
output:
2 20 0
result:
ok 3 number(s): "2 20 0"
Test #2:
score: -100
Runtime Error
input:
1060 19 12 0 UDLDDUUUUDDDLLRDUDUURULUUUDRDUDRDRLRLRLULULLLDLDDRLUUUURUUUDDRLLRUUUDULURUULLRDRLRDDURDUUURRRLURLRUULRRUDURDLUUURDLURDDLUUURDDRLLURRDLRUDLRDRLLRRDRDDLDRURRRLUDULLLRUUDLRRURRDLLRRRDLLRDDDLRLRURURDDDL 11 1 0 UR 3 18 33 UDRLR 17 11 132 RLDRDLDRUU 6 10 13 UULUDDLRDLUUDLDD 1 15 0 D 6 20 50 D...