noip模拟78

A. F

签到题，随便写.

A_code

#include<bits/stdc++.h>
using namespace std;
namespace BSS {
	#define ll long long int
	#define ull unsigned ll
	#define lf long double
	#define lbt(x) (x&(-x))
	#define mp(x,y) make_pair(x,y)
	#define lb lower_bound 
	#define ub upper_bound
	#define Fill(x,y) memset(x,y,sizeof x)
	#define Copy(x,y) memcpy(x,y,sizeof x)
	#define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
	inline ll read() {
		ll w=0; bool cit=1; char ch;
		while(!isdigit(ch=getchar())) if(ch=='-') cit=0; 
		while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar();
		return cit?w:-w;
	}
} using namespace BSS;

#define fre(x,y) freopen(#x,"r",stdin),freopen(#y,"w",stdout)
const ll N=2e3+21,M=32;

ll m,n,lg2,cnt,maxval;
ll a[N],b[N],bin[N];
unordered_map<ll,ll> vis;
signed main(){
	fre(f.in,f.out);
	n=read(); ll x,y,res,flag;
	for(ll i=1;i<=n;i++) a[i]=read(),maxval=max(maxval,a[i]);
	for(ll i=1;i<=n;i++) b[i]=read(),maxval=max(maxval,b[i]);
	lg2=log2(maxval)+1,cnt=0;
	for(ll i=0;i<=lg2;i++){
		x=0,y=0;
		for(ll j=1;j<=n;j++) if((1ll<<i)&a[j]) x++;
		for(ll j=1;j<=n;j++) if((1ll<<i)&b[j]) y++;
		if(x!=y and x+y!=n) puts("0"),exit(0);
	}
	for(ll i=1;i<=n;i++){
		res=a[1]^b[i],flag=1;
		for(ll j=1;j<=n;j++) vis[b[j]]=0;
		for(ll j=1;j<=n;j++) vis[b[j]]++;
		for(ll j=1;j<=n;j++){
			if(vis[a[j]^res]) vis[a[j]^res]--;
			else { flag=0; break; }
		}
		if(flag) bin[++cnt]=res;
	}
	sort(bin+1,bin+1+cnt),printf("%lld\n",cnt);
	for(ll i=1;i<=cnt;i++) printf("%lld\n",bin[i]);
	exit(0);
}

B. S

交换次数可以被联想成逆序对.

我们可以回想归并排序的求解过程，发现有序的依旧有序，无序的才会被交换.

本题中类比过来就是，有序的就是那些相同颜色的，也就是相同颜色的不会换位置.

于是可以写出 \(dp_{i,j,k,c}\) 表示 \(R\) 填了 \(i\) 个，\(G\) 填了 \(j\) 个，\(Y\) 填了 \(k\) 个，序列最后的字母是 \(c\) 的最小交换次数.

B_code

#include<bits/stdc++.h>
using namespace std;
namespace BSS {
	#define ll int
	#define ull unsigned ll
	#define lf double
	#define lbt(x) (x&(-x))
	#define mp(x,y) make_pair(x,y)
	#define lb lower_bound 
	#define ub upper_bound
	#define Fill(x,y) memset(x,y,sizeof x)
	#define Copy(x,y) memcpy(x,y,sizeof x)
	#define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
	inline ll read() {
		ll w=0; bool cit=1; char ch;
		while(!isdigit(ch=getchar())) if(ch=='-') cit=0; 
		while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar();
		return cit?w:-w;
	}
} using namespace BSS;

const ll N=405,M=205;

ll m,n,ans;
ll c[N],cnt[3];
ll pos[3][N],pre[3][N];
ll dp[M][M][M][3];
inline void ckmin(ll &x,ll y){ x=min(x,y); }
signed main(){
	File(s);
	n=read(); char ch[N]; scanf("%s",ch+1);
	for(ll i=1;i<=n;i++){
		if(ch[i]=='R') c[i]=0; if(ch[i]=='G') c[i]=1; if(ch[i]=='Y') c[i]=2;
		pos[c[i]][++cnt[c[i]]]=i; pre[0][i]=cnt[0],pre[1][i]=cnt[1],pre[2][i]=cnt[2];
		// cout<<pre[0][i]<<' '<<pre[1][i]<<' '<<pre[2][i]<<endl;	
	}
	if(cnt[0]>(n+1>>1) or cnt[1]>(n+1>>1) or cnt[2]>(n+1>>1)) puts("-1"),exit(0);
	Fill(dp,0x3f),Fill(dp[0][0][0],0); ll now,nxt,res;
	for(ll i=0;i<=cnt[0];i++){
		for(ll j=0;j<=cnt[1];j++)
			for(ll k=0;k<=cnt[2];k++){
				// cout<<i<<' '<<j<<' '<<k<<' '<<dp[i][j][k][0]<<' '<<dp[i][j][k][1]<<" "<<dp[i][j][k][2]	<<endl;
				nxt=pos[0][i+1],res=max(0,pre[1][nxt]-pre[1][pos[1][j]])+max(0,pre[2][nxt]-pre[2][pos[2][k]]);
				ckmin(dp[i+1][j][k][0],min(dp[i][j][k][1],dp[i][j][k][2])+res);
				nxt=pos[1][j+1],res=max(0,pre[0][nxt]-pre[0][pos[0][i]])+max(0,pre[2][nxt]-pre[2][pos[2][k]]);
				ckmin(dp[i][j+1][k][1],min(dp[i][j][k][0],dp[i][j][k][2])+res);
				nxt=pos[2][k+1],res=max(0,pre[0][nxt]-pre[0][pos[0][i]])+max(0,pre[1][nxt]-pre[1][pos[1][j]]);
				ckmin(dp[i][j][k+1][2],min(dp[i][j][k][0],dp[i][j][k][1])+res);
			}
	}
	ans=min(dp[cnt[0]][cnt[1]][cnt[2]][0],dp[cnt[0]][cnt[1]][cnt[2]][1]);
	ans=min(dp[cnt[0]][cnt[1]][cnt[2]][2],ans);
	printf("%d\n",ans),exit(0);
}

C. Y

D. O

扫描线，不会写，先鸽了.