noip模拟61
A. 交通
考虑转化问题.
把边之间的禁止关系转化成点之间的.
还有一种思路.
发现每个边的出边和入边只能选一个.
所以可以选择固定一个点的出边/入边,然后删边判定就行了.
A_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long 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 res=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
return cit?res:-res;
}
} using namespace BSS;
const ll N=3e6+21,mod=998244353;
ll m,n,ans;
ll fa[N],vis[N];
struct Graph{
ll ts; ll head[N];
struct I { ll u,v,col,nxt; } e[N<<1];
inline void add(ll u,ll v,ll opt){
e[++ts].u=u,e[ts].v=v,e[ts].nxt=head[u],
head[u]=ts,e[ts].col=opt;
}
}A,B;
inline ll find(ll x){ return x==fa[x] ? x : fa[x]=find(fa[x]); }
inline ll ksm(ll a,ll b,ll c){
a%=c; ll res=1;
for(;b;b>>=1,a=a*a%c) if(b&1) res=res*a%c;
return res%c;
}
signed main(){
File(a);
n=read(); ll u,v,a,b,c,d; A.ts=1;
for(int i=1;i<=2*n;i++) u=read(),v=read(),A.add(u,v,1),A.add(v,u,0);
for(int u=1;u<=n;u++){
a=A.head[u],b=A.e[a].nxt,c=A.e[b].nxt,d=A.e[c].nxt;
if(A.e[a].col==A.e[b].col) B.add(a-(a&1),b-(b&1),0),B.add(c-(c&1),d-(d&1),0);
else if(A.e[a].col==A.e[c].col) B.add(a-(a&1),c-(c&1),0),B.add(b-(b&1),d-(d&1),0);
else if(A.e[a].col==A.e[d].col) B.add(a-(a&1),d-(d&1),0),B.add(b-(b&1),c-(c&1),0);
}
for(int i=2;i<=A.ts;i+=2) fa[i]=i;
for(int i=1;i<=B.ts;i++){
u=B.e[i].u,v=B.e[i].v;
if(find(u)!=find(v)) fa[find(u)]=find(v);
}
for(int i=2;i<=A.ts;i+=2) if(!vis[find(i)]) ans++,vis[find(i)]=1;
printf("%lld\n",ksm(2,ans,mod));
exit(0);
}
B. 小P的单调数列
一开始就应该推性质而不是莽干的.
做不出来的时候思维就应该发散一下.
由于越取平均数答案越小,其实序列有一个最多只有两段的性质.
也就是说答案要么只有单增的一段,要么就是一段单增,一段单减.
B_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long 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 res=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
return cit?res:-res;
}
} using namespace BSS;
const ll N=1e5+21;
lf ans;
ll m,n,cnt;
ll val[N],lsh[N],ans1[N],ans2[N];
struct Bit_Array{
ll res; ll c[N<<2];
inline void update(ll x,ll w){
for(;x<=n;x+=lbt(x)) c[x]=max(c[x],w);
}
inline ll query(ll x){
res=0; for(;x>0;x&=(x-1)) res=max(res,c[x]); return res;
}
inline void clr(){ Fill(c,0); }
}bit;
signed main(){
File(b);
n=read();
for(int i=1;i<=n;i++) lsh[i]=val[i]=read();
sort(lsh+1,lsh+1+n),cnt=unique(lsh+1,lsh+1+n)-lsh-1;
for(int i=1;i<=n;i++){
val[i]=lb(lsh+1,lsh+1+cnt,val[i])-lsh;
ans1[i]=bit.query(val[i]-1)+lsh[val[i]];
bit.update(val[i],ans1[i]);
}
bit.clr();
for(int i=n;i>=1;i--){
ans2[i]=bit.query(val[i]-1)+lsh[val[i]];
bit.update(val[i],ans2[i]);
}
for(int i=1;i<=n;i++) ans1[i]=max(ans1[i],ans1[i-1]);
for(int i=n;i>=1;i--) ans2[i]=max(ans2[i],ans2[i+1]);
for(int i=1;i<=n;i++){
ans=max(ans,1.0*ans1[i]);
ans=max(ans,1.0*(ans1[i]+ans2[i+1])/2);
}
printf("%.3lf\n",ans);
exit(0);
}
C. 矩阵
很显然是构造题.
由于两点之间即可确定一条线.
所以可以转化成只让前两行和前两列被消成 \(0\).
C_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long 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 res=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
return cit?res:-res;
}
} using namespace BSS;
const ll N=1e3+21;
ll m,n,cntx,cnty,cntz;
ll x[N],y[N],z[N*3];
ll w[N][N];
signed main(){
File(c);
n=read(),m=read();
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
w[i][j]=read();
}
}
x[1]=w[2][2]-w[1][1],z[N]=-w[2][2];
for(int i=2;i<=n;i++) z[N+1-i]+=-x[i]-w[i][1],x[i+1]+=-(z[N+1-i]+w[i+1][2]);
for(int j=2;j<=m;j++) z[N+j-1]+=-y[j]-x[1]-w[1][j],y[j+1]+=-(z[N+j-1]+w[2][j+1]);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++)
if(w[i][j]+x[i]+y[j]+z[j-i+N]) puts("-1"),exit(0);
}
cout<<n+m+(m-1-(1-n)+1)<<endl;
for(int i=1;i<=n;i++) cout<<1<<' '<<i<<' '<<x[i]<<endl;
for(int j=1;j<=m;j++) cout<<2<<' '<<j<<' '<<y[j]<<endl;
for(int i=1-n;i<=m-1;i++) cout<<3<<' '<<i<<' '<<z[N+i]<<endl;
exit(0);
}
D. 花瓶
没见过二维斜率优化,枯了.
考场上想了斜率优化,但是写的是一维的,发现这个东西很离奇,于是就死了.
不应该拘束于已经学过的东西.
D_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS {
#define ll long long 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 res=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
return cit?res:-res;
}
} using namespace BSS;
const ll N=5e3+21;
ll m,n,ans;
ll val[N],stk[N],sum[N];
ll dp[N][N];
struct I { ll pre,id; } p[N];
inline bool comp(I i,I j){ return i.pre==j.pre ? i.id<j.id : i.pre<j.pre; }
inline ll caly(ll j,ll k){ return dp[j][k]; }
inline lf slope(ll j,ll ka,ll kb){ return sum[ka]==sum[kb] ? 1e15 : 1.0*(caly(j,ka)-caly(j,kb))/(sum[ka]-sum[kb]); }
signed main(){
File(d);
n=read(); ll l,r,id;
for(int i=1;i<=n;i++){
val[i]=read(),p[i].pre=p[i-1].pre+val[i],p[i].id=i;
sum[i]=p[i].pre;
}
sort(p+1,p+1+n,comp);
for(int j=1;j<=n;j++){
l=1,r=0;
for(int k=1;k<=n;k++){
id=p[k].id; if(id>=j) continue;
while(l<r and slope(j,stk[r],stk[r-1])<=slope(j,id,stk[r])){
r--;
}
stk[++r]=id;
}
for(int i=n;i>=1;i--){
id=p[i].id; if(id<=j) continue;
while(l<r and (slope(j,stk[l+1],stk[l])>=(sum[id]-sum[j]))){
l++;
}
dp[id][j]=max(caly(j,stk[l])+(sum[id]-sum[j])*(sum[j]-sum[stk[l]]),(sum[id]-sum[j])*sum[j]);
}
}
for(int i=1;i<=n;i++){
ans=max(ans,dp[n][i]);
}
printf("%lld\n",ans);
exit(0);
}
