GXOI/GZOI2019部分题解

D1T1:与或和

对每位处理，问题变成所有内部不包含0/1的矩阵的个数，单调栈维护即可。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,mod=1e9+;
 int top,sum,sm,n,mx,a[N][N],b[N][N],st[N],sz[N],l[N][N];

 int work(int w,int op){
     rep(i,,n) rep(j,,n)
         if((a[i][j]&(<<w))) b[i][j]=op; else b[i][j]=op^;
     rep(i,,n){
         l[i][n]=b[i][n];
         for(int j=n-; j; j--)
             if(!b[i][j]) l[i][j]=; else l[i][j]=l[i][j+]+;
     }
     int tmp=;
     rep(j,,n){
         top=,sum=;
         rep(i,,n){
             int now=;
             while (top && st[top]>=l[i][j])
                 sum-=st[top]*sz[top],now+=sz[top],top--;
             st[++top]=l[i][j]; sz[top]=now+;
             sum+=st[top]*sz[top]; tmp=(tmp+sum)%mod;
         }
     }
     return tmp;
 }

 int main(){
     freopen("andorsum.in","r",stdin);
     freopen("andorsum.out","w",stdout);
     scanf("%d",&n);
     rep(i,,n) rep(j,,n) scanf("%d",&a[i][j]),mx=max(mx,a[i][j]);
     rep(i,,n) rep(j,,n) sm=(sm+1ll*i*j)%mod;
     int ans1=,ans2=;
     rep(w,,){
         if ((1ll<<w)>mx) break;
         int t=(1ll<<w)%mod;
         ans1=(ans1+1ll*t*work(w,)%mod)%mod;
         ans2=(ans2+1ll*t*(sm-work(w,)+mod)%mod)%mod;
     }
     printf("%d %d\n",ans1,ans2);
     return ;
 }

andorsum

D1T3:特技飞行

被观察到的特技次数是固定的，先曼哈顿转切比雪夫然后KDT统计即可。

然后发现，若所有特技全部选择对向交换，一定是符合要求的，于是得到了两个极值中的一个。

接下来要让对向交换的次数尽量小，也就是给两个排列，让第一个经过最小的交换次数变成第二个。

找出置换环，答案就是n-置换环数。

 #include<set>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=;
 const double eps=1e-,inf=1e13;
 ll A,B,C;
 int n,T,tot,st,ed,x,y,rr,y0[N],y1[N],id[N],vis[N],b[N],cov[N];
 set<pair<int,int> >S;
 set<pair<int,int> >::iterator it;

 struct P{ double v[],mn[],mx[]; int ls,rs; }v[N],t;
 bool cmpx(const P &a,const P &b){ return a.v[]<b.v[]; }
 bool cmpy(const P &a,const P &b){ return a.v[]<b.v[]; }
 bool cmp(int a,int b){ return y1[a]<y1[b]; }

 double Abs(double x){ return x< ? -x : x; }

 void upd(int x){
     rep(i,,){
         v[x].mn[i]=min(v[x].v[i],min(v[v[x].ls].mn[i],v[v[x].rs].mn[i]));
         v[x].mx[i]=max(v[x].v[i],max(v[v[x].ls].mx[i],v[v[x].rs].mx[i]));
     }
 }

 int build(int L,int R,int k){
     if (L>R) return ;
     int mid=(L+R)>>,x=mid;
     nth_element(v+L,v+mid,v+R+,k?cmpy:cmpx);
     v[x].ls=build(L,mid-,k^); v[x].rs=build(mid+,R,k^);
     upd(x); return x;
 }

 void mdf(int x){
     if (!x || b[x] || t.v[]+rr<v[x].mn[] || t.v[]-rr>v[x].mx[] || t.v[]+rr<v[x].mn[] || t.v[]-rr>v[x].mx[]) return;
     double s=;
     rep(i,,) s=max(s,max(Abs(t.v[i]-v[x].mn[i]),Abs(t.v[i]-v[x].mx[i])));
     if (s-eps<rr){ b[x]=; return; }
     if (!cov[x] && max(Abs(v[x].v[]-t.v[]),Abs(v[x].v[]-t.v[]))-eps<rr) cov[x]=;
     mdf(v[x].ls); mdf(v[x].rs);
 }

 int dfs(int x){
     if (b[x]) cov[x]=cov[v[x].ls]=cov[v[x].rs]=b[v[x].ls]=b[v[x].rs]=;
     int res=cov[x];
     if (v[x].ls) res+=dfs(v[x].ls);
     if (v[x].rs) res+=dfs(v[x].rs);
     return res;
 }

 int main(){
     freopen("aerobatics.in","r",stdin);
     freopen("aerobatics.out","w",stdout);
     v[].mn[]=v[].mn[]=inf; v[].mx[]=v[].mx[]=-inf;
     scanf("%d%lld%lld%lld%d%d",&n,&A,&B,&C,&st,&ed);
     rep(i,,n) scanf("%d",&y0[i]);
     rep(i,,n) scanf("%d",&y1[i]);
     for (int i=n; i; i--){
         for (it=S.begin(); it!=S.end() && (it->first<y1[i]); it++){
             int j=it->second;
             double t=(double)(y0[j]-y0[i])/(y1[i]-y1[j]);
             double x=(t*ed+st)/(t+),y=(t*y1[j]+y0[j])/(t+);
             v[++tot].v[]=x+y; v[tot].v[]=x-y;
         }
         S.insert(pair<int,int>(y1[i],i));
     }
     int rt=build(,tot,);
     for (scanf("%d",&T); T--; )
         scanf("%d%d%d",&x,&y,&rr),t.v[]=x+y,t.v[]=x-y,mdf(rt);
     ll ans=dfs(rt)*C,ans1=ans+tot*A,ans2=ans+tot*A;
     rep(i,,n) id[i]=i;
     sort(id+,id+n+,cmp);
     ll x=n;
     rep(i,,n) if (!vis[i]){
         x--;
         for (int j=i; !vis[j]; j=id[j]) vis[j]=;
     }
     ans2+=(tot-x)*(B-A);
     printf("%lld %lld\n",min(ans1,ans2),max(ans1,ans2));
     return ;
 }

aerobatics

D2T2:旅行者

方法一：二进制分组，对每个二进制位，将所有编号在这一位为0的关键点放在一边，为1的放在另一边跑最短路。这样任意两点间最短路都一定在某次中被考虑到。O(nlog^2n)

 #include<queue>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 #define For(i,x) for (int i=h[x],k; i; i=nxt[i])
 using namespace std;

 const int N=,inf=1e9;
 int n,m,k,T,u,v,w,ans,cnt,s[N],b[N],dis[N],fr[N],h[N],to[N],nxt[N],val[N];
 struct P{ int x,d; };
 bool operator <(const P &a,const P &b){ return a.d>b.d; }
 priority_queue<P>Q;

 void add(int u,int v,int w){ to[++cnt]=v; val[cnt]=w; nxt[cnt]=h[u]; h[u]=cnt; }

 void dij(){
     while (!Q.empty()){
         int x=Q.top().x; Q.pop();
         if (b[x]) continue;
         b[x]=;
         For(i,x) if (dis[k=to[i]]>dis[x]+val[i])
             dis[k]=dis[x]+val[i],fr[k]=fr[x],Q.push((P){k,dis[k]});
     }
 }

 int main(){
     freopen("tourist.in","r",stdin);
     freopen("tourist.out","w",stdout);
     for (scanf("%d",&T); T--; ){
         cnt=; ans=inf; rep(i,,n) h[i]=;
         scanf("%d%d%d",&n,&m,&k);
         rep(i,,m) scanf("%d%d%d",&u,&v,&w),add(u,v,w);
         rep(i,,k) scanf("%d",&s[i]);
         for (int t=; t<=k; t<<=){
             rep(i,,n) dis[i]=inf,b[i]=;
             while (!Q.empty()) Q.pop();
             rep(i,,k) if (i&t) dis[s[i]]=,Q.push((P){s[i],}),fr[s[i]]=s[i];
             dij();
             rep(i,,k) if (!(i&t)) ans=min(ans,dis[s[i]]);
             rep(i,,n) dis[i]=inf,b[i]=;
             while (!Q.empty()) Q.pop();
             rep(i,,k) if (!(i&t)) dis[s[i]]=,Q.push((P){s[i],}),fr[s[i]]=s[i];
             dij();
             rep(i,,k) if (i&t) ans=min(ans,dis[s[i]]);
         }
         printf("%d\n",ans);
     }
     return ;
 }

方法一

方法二：对于一条边，它对答案的贡献是a[u]+w+b[v]，其中a是所有关键点到它的最短距离，b是它到所有关键点的最短距离。

注意若a和b来自同一个关键点的时候忽略这条边，显然任意两个关键点之间的最短路上总有一条边不会被忽略。O(nlogn)

 #include<queue>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 #define For(i,x) for (int i=h[x],k; i; i=nxt[i])
 using namespace std;

 const int N=,inf=1e9;
 int n,m,k,T,u,v,w,ans,s[N],b[N];
 struct P{ int x,d; };
 bool operator <(const P &a,const P &b){ return a.d>b.d; }
 priority_queue<P>Q;
 struct E{ int u,v,w; }e[N];

 struct Gr{
     int cnt,dis[N],fr[N],h[N],to[N],nxt[N],val[N];
     void add(int u,int v,int w){ to[++cnt]=v; val[cnt]=w; nxt[cnt]=h[u]; h[u]=cnt; }

     void dij(){
         rep(i,,n) dis[i]=inf,b[i]=;
         while (!Q.empty()) Q.pop();
         rep(i,,k) dis[s[i]]=,Q.push((P){s[i],}),fr[s[i]]=s[i];
         while (!Q.empty()){
             int x=Q.top().x; Q.pop();
             if (b[x]) continue;
             b[x]=;
             For(i,x) if (dis[k=to[i]]>dis[x]+val[i])
                 dis[k]=dis[x]+val[i],fr[k]=fr[x],Q.push((P){k,dis[k]});
         }
     }
 }G1,G2;

 void init(){
     G1.cnt=G2.cnt=; ans=inf;
     memset(G1.h,,sizeof(G1.h)); memset(G2.h,,sizeof(G2.h));
 }

 int main(){
     freopen("tourist.in","r",stdin);
     freopen("tourist.out","w",stdout);
     for (scanf("%d",&T); T--; ){
         init(); scanf("%d%d%d",&n,&m,&k);
         rep(i,,m) scanf("%d%d%d",&u,&v,&w),G1.add(u,v,w),G2.add(v,u,w),e[i]=(E){u,v,w};
         rep(i,,k) scanf("%d",&s[i]);
         G1.dij(); G2.dij();
         rep(i,,m) if (G1.fr[e[i].u]!=G2.fr[e[i].v]) ans=min(ans,G1.dis[e[i].u]+e[i].w+G2.dis[e[i].v]);
         printf("%d\n",ans);
     }
     return ;
 }

方法二