A. Space Navigation
思路：分别统计往px和py方向的步数，看看能不能分别大于等于px和py。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int main()
{
int kase;
cin>>kase;
while(kase--)
{
ll x = read(), y = read();
string s;
cin>>s;
ll xx1 = 0, yy1 = 0, xx2 = 0, yy2 = 0;
for(int i=0; i<s.size(); i++)
{
if(s[i]=='L') xx2--;
else if(s[i]=='R') xx1++;
else if(s[i]=='U') yy1 ++;
else yy2++;
}
int flag = 0;
int flag1 = 0;
if(y>=0&&yy1>=y) flag = 1;
else if(y<=0&&abs(yy2)>=abs(y)) flag = 1;
if(x>=0&&xx1>=x) flag1 = 1;
else if(x<=0&&abs(xx2)>=abs(x)) flag1 = 1;
puts(flag&&flag1?"YES":"NO");
}
return 0;
}
B. New Colony
思路：数据量小，直接模拟就行。最坏的情况全部走完也不需要很多k。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
int main()
{
int kase;
cin>>kase;
while(kase--)
{
ll n = read(), k = read();
rep(i,1,n) a[i] = read();
int pos = 0;
int flag = 0;
while(k>0)
{
k--;
rep(i,1,n)
{
if(i==n)
{
flag = 1;
break;
}
if(a[i]>=a[i+1]) continue;
pos = i;
a[i]++;
break;
}
if(flag) break;
}
if(k||flag) cout<<-1<<endl;
else cout<<pos<<endl;
}
return 0;
}
C. Fence Painting
思路：策略如下：
定义要改的数为(a[i]!=b[i])的。
1.我遍历到当前的c[i],如果这个c[i]就是我要改的数，那优先改。
2.否则，如果当前这个数不是我要改的数，但是它是某个a[i]==b[i]产生的“不用改”，那就可以在对应位置修改。（这个是在不满足条件1的情况下才采取的下策）。
3.若还不满足上述条件，那就说明当前还找不到具体的位置放，那就先存起来，一旦后面遍历到的c[i]满足上述((1) (2))之一，就可以让这些存起来的数放到他们的位置，反正会被覆盖掉。
4.详见代码注释吧。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
ll b[maxn];
ll c[maxn];
ll Ans[maxn];
int main()
{
int kase;
cin>>kase;
while(kase--)
{
ll n = read(), m = read();
map<ll,stack<ll> > Map; // Map存放需要改变的位置
map<ll,ll> Pos; //Pos记录一下每个b[i]对应的i，多个b[i]取最后那个
rep(i,1,n) a[i] = read();
rep(i,1,n)
{
b[i] = read();
if(a[i]!=b[i]) //只存放需要改变的位置
Map[b[i]].push(i);
Pos[b[i]] = i; //记录b[i]位置
}
rep(i,1,m) c[i] = read();
stack<ll > pre; //pre记录一下前面遇到没地方放的数
int flag = 1;
rep(i,1,m)
{
if(Map[c[i]].size()) //如果当前c[i]有对应需要改变的 ，就优先改
{
ll cur = Map[c[i]].top();
Map[c[i]].pop();
Ans[i] = cur;
a[cur] = b[cur];
while(pre.size()) // 同时把前面没地方放的那些数放到当前这个位置，意思就是前面随便改我这个位置，利用当前c[i]我又改回来
{
ll id = pre.top();
pre.pop();
Ans[id] = cur;
}
}
else //若没有一定要改的位置
{
if(Pos[c[i]]&&pre.size()) //先看看这个数存不存在(Pos[c[i]])，若同时前面有数放，则同上
{
while(pre.size())
{
ll id = pre.top();
pre.pop();
Ans[id] = Pos[c[i]];
Ans[i] = Pos[c[i]];
a[Pos[c[i]]] = b[Pos[c[i]]];
}
}
else if(Pos[c[i]]) Ans[i] = Pos[c[i]]; //前面没有数的但可以放，就放当前
else pre.push(i); //其他就没办法了，只能放进pre，让后面的c[j]来完成
}
}
rep(i,1,n) if(a[i]!=b[i]) flag = 0;
if(pre.size()) flag = 0;
if(flag)
{
cout<<"YES"<<endl;
rep(i,1,m) cout<<Ans[i]<<' '; cout<<endl;
}
else cout<<"NO"<<endl;
}
return 0;
}