A. Odd Palindrome
所有迴文子串長度都是奇數等價於不存在長度為$2$的偶迴文子串，即相鄰兩個字元都不同。
#include<cstdio>
#include<cstring>
char a[1111111];int n,i;
int main(){
scanf("%s",a+1);
n=strlen(a+1);
for(i=1;i<n;i++)if(a[i]==a[i+1])return puts("Or not."),0;
puts("Odd.");
}
B. Enlarging Enthusiasm
注意到方案數不超過$(n-1)\times (n-1)!$，爆搜出所有可行方案即可，需要大量常數優化。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=15;
int n,m,i,a[N],f[(1<<12)+5],ans;
int cnt[(1<<12)+5];
int w[N];
void dfs(int S,int mx,int q,int x){
if(!S){
ans++;
return;
}
mx++;
if((S&-S)==S){
int j=f[S];
int nowq=mx-j>q?mx-j:q;
if(x>=nowq)ans++;
return;
}
int U=S;
while(U){
int i=U&-U,j=f[i];
int nowq=mx-j>q?mx-j:q;
if(x>=nowq)dfs(S^i,j+nowq,nowq,x-nowq);
U^=i;
}
}
void gao(int S,int mx,int q,int x){
if(!S){
ans++;
return;
}
mx++;
for(int U=S;U;U-=U&-U){
int i=U&-U;
if(i==(1<<(n-1)))continue;
int nowq=mx-f[i];
if(nowq<q)nowq=q;
if(x>=nowq)dfs(S^i,f[i]+nowq,nowq,x-nowq);
}
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<1<<n;i++)cnt[i]=cnt[i>>1]+(i&1);
for(w[0]=i=1;i<=n;i++)w[i]=w[i-1]*i;
for(i=0;i<n;i++)scanf("%d",&a[i]);
sort(a,a+n);
for(i=0;i<n;i++)f[1<<i]=a[i];
gao((1<<n)-1,a[n-1],0,m);
printf("%d",ans);
}
/*
12 700
1 2 3 4 5 6 7 8 9 10 11 12
*/
C. Fear Factoring
考慮每個約數的貢獻，那麼分段等差數列求和即可。
時間複雜度$O(\sqrt{b})$。
#include<cstdio>
typedef long long ll;
ll cal(ll n){
ll i,j;
ll ans=0;
for(i=1;i<=n;i=j+1){
j=n/(n/i);
ans+=(i+j)*(j-i+1)*(n/i);
}
return ans;
}
int main(){
ll a,b;
scanf("%lld%lld",&a,&b);
printf("%lld",(cal(b)-cal(a-1))/2);
}
D. Rainbow Roads
對於每個點，若其有超過一條同色的邊，那麼對應子樹都不是Good點，差分字首和打標記即可。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef pair<int,int>P;
const int N=300010;
int n,i,j,k,x,y,z,g[N],v[N],w[N],nxt[N],ed;
int f[N],dfn,st[N],en[N];
int m,ans,s[N];
P q[N];
inline void add(int x,int y,int z){
v[++ed]=y;
w[ed]=z;
nxt[ed]=g[x];
g[x]=ed;
}
void dfs(int x,int y){
st[x]=++dfn;
f[x]=y;
for(int i=g[x];i;i=nxt[i])if(v[i]!=y)dfs(v[i],x);
en[x]=dfn;
}
int main(){
scanf("%d",&n);
for(i=1;i<n;i++){
scanf("%d%d%d",&x,&y,&z);
add(x,y,z),add(y,x,z);
}
dfs(1,0);
for(i=1;i<=n;i++){
m=0;
for(j=g[i];j;j=nxt[j])q[++m]=P(w[j],v[j]);
sort(q+1,q+m+1);
for(j=1;j<=m;j=k){
for(k=j;k<=m&&q[j].first==q[k].first;k++);
if(k>j+1){
for(x=j;x<k;x++){
y=q[x].second;
if(y==f[i]){
s[1]++;
s[st[i]]--;
s[en[i]+1]++;
}else{
s[st[y]]++;
s[en[y]+1]--;
}
}
}
}
}
for(i=1;i<=n;i++)s[i]+=s[i-1];
for(i=1;i<=n;i++)if(!s[st[i]])ans++;
printf("%d\n",ans);
for(i=1;i<=n;i++)if(!s[st[i]])printf("%d\n",i);
}
E. Straight Shot
二分水平分速度，檢查最終是否走到了$(x,0)$。
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=111;
int n,i;
double X,V,dx,dy;
double lim,goal=1e100,tmp,L,R,MID,l[N],r[N],v[N];
double cal(double dy){
double dx=sqrt(V*V-dy*dy);
double x=0,y=0;
for(int i=1;i<=n;i++){
y+=dy*(l[i]-x);
y+=(dy+v[i])*(r[i]-l[i]);
x=r[i];
}
y+=dy*(X-x);
tmp=X/dx;
return y/dx;
}
int main(){
scanf("%d%lf%lf",&n,&X,&V);
for(i=1;i<=n;i++)scanf("%lf%lf%lf",&l[i],&r[i],&v[i]);
L=-V,R=V;
for(int _=1000;_;_--){
MID=(L+R)/2;
if(cal(MID)<0)L=MID;else R=MID;
}
lim=X/V*2;
L=(L+R)/2;
if(fabs(cal(L))<1e-8){
cal(L);
goal=tmp;
}
if(goal>lim+1e-8)puts("Too hard");
else printf("%.3f",goal);
}
F. Distinct Distances
答案點只可能取在每個點、兩個點的中點，以及兩對點垂直平分線的交點。
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=50;
const double eps=1e-9;
inline int sgn(double x){
if(x>eps)return 1;
if(x<-eps)return -1;
return 0;
}
struct P{
double x,y;
P(){}
P(double _x,double _y){x=_x,y=_y;}
P operator+(P b){return P(x+b.x,y+b.y);}
P operator-(P b){return P(x-b.x,y-b.y);}
P operator*(double b){return P(x*b,y*b);}
P operator/(double b){return P(x/b,y/b);}
void read(){scanf("%lf%lf",&x,&y);}
double len(){return x*x+y*y;}
double dis(const P&b){return (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y);}
bool operator==(P b){return !sgn(x-b.x)&&!sgn(y-b.y);}
P rot90(){return P(-y,x);}
}a[N],e[N*N][2];
int n,m,i,j,ans=N;
double d[N];
inline void solve(P O){
//printf("%.10f %.10f\n",O.x,O.y);
int i;
for(i=1;i<=n;i++)d[i]=O.dis(a[i]);
sort(d+1,d+n+1);
int t=1;
for(i=2;i<=n&&t<ans;i++)if(d[i]>d[i-1]+eps)t++;
if(t<ans)ans=t;
}
inline void makeline(P a,P b){
if(a==b)return;
P c=(a+b)/2.0;
solve(c);
b=b-a;
e[++m][0]=c;
e[m][1]=c+b.rot90();
//printf("%.5f %.5f %.5f %.5f\n",e[m][0].x,e[m][0].y,e[m][1].x,e[m][1].y);
}
inline double cross(P a,P b){
return a.x*b.y-a.y*b.x;
}
inline void gao(P a,P b,P p,P q){
double U=cross(p-a,q-p),D=cross(b-a,q-p);
if(!sgn(D))return;
//puts("!");
solve(a+(b-a)*(U/D));
}
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++)a[i].read();
for(i=1;i<=n;i++)solve(a[i]);
for(i=1;i<=n;i++)for(j=1;j<i;j++){
makeline(a[i],a[j]);
}
//makeline(a[4],a[5]);
//makeline(a[5],a[1]);
for(i=1;i<=m;i++)for(j=1;j<i;j++)gao(e[i][0],e[i][1],e[j][0],e[j][1]);
printf("%d",ans);
}
/*
6
0 -5
1 0
-1 0
2 3
3 2
-3 0
*/
G. Security Badge
對區間離散化，那麼只需要檢查$O(m)$個人是否可行，每次暴力Floodfill判斷即可。
時間複雜度$O(m(n+m))$。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=10010;
int n,m,k,S,T,i,j,cnt,q[N],K;
int g[N],ed,v[N],vc[N],vd[N],nxt[N],vis[N],ans;
inline void add(int x,int y,int c,int d){
v[++ed]=y;
vc[ed]=c;
vd[ed]=d;
nxt[ed]=g[x];
g[x]=ed;
}
void dfs(int x){
if(vis[x])return;
vis[x]=1;
for(int i=g[x];i;i=nxt[i])if(vc[i]<=K&&K<=vd[i])dfs(v[i]);
}
int main(){
scanf("%d%d%d%d%d",&n,&m,&k,&S,&T);
for(i=1;i<=m;i++){
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
add(a,b,c,d);
q[++cnt]=c-1;
q[++cnt]=d;
}
sort(q+1,q+cnt+1);
for(i=2;i<=cnt;i++)if(q[i]!=q[i-1]){
K=q[i];
for(j=1;j<=n;j++)vis[j]=0;
dfs(S);
if(vis[T])ans+=q[i]-q[i-1];
}
printf("%d",ans);
}
H. Avoiding Airports
將每架飛機拆成上飛機和下飛機兩個事件，並按照時間順序依次考慮。
設$dp[x]$表示最後乘坐第$x$架飛機的最小代價，那麼轉移是經典斜率優化形式，對於每個點用單調佇列維護凸殼即可。
時間複雜度$O(m\log m)$。
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int N=200010;
const ll inf=1LL<<60;
int n,m,i,e[N][4],cnt;
struct E{
int x,y;
E(){}
E(int _x,int _y){x=_x,y=_y;}
}w[N<<1];
ll dp[N],ans=inf;
struct Line{
ll k,b;
Line(){}
Line(ll _k,ll _b){k=_k,b=_b;}
ll f(ll x){return k*x+b;}
};
vector<Line>g[N];
int head[N],tail[N],deg[N];
inline bool cmp(const E&a,const E&b){
if(a.x!=b.x)return a.x<b.x;
return a.y<b.y;
}
inline double pos(const Line&a,const Line&b){
return 1.0*(a.b-b.b)/(b.k-a.k);
}
inline void ins(int o,ll k,ll b){
if(b>=inf)return;
Line now(k,b);
int&h=head[o],&t=tail[o];
if(h<=t){
if(g[o][t].k==k){
if(g[o][t].b<=b)return;
t--;
}
}
while(h<t&&pos(g[o][t-1],g[o][t])>=pos(g[o][t],now))t--;
g[o][++t]=now;
}
inline ll cal(int o,ll x){
int&h=head[o],&t=tail[o];
if(h>t)return inf;
while(h<t&&g[o][h].f(x)>g[o][h+1].f(x))h++;
return g[o][h].f(x);
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++){
scanf("%d%d%d%d",&e[i][0],&e[i][1],&e[i][2],&e[i][3]);
deg[e[i][1]]++;
w[++cnt]=E(e[i][2],-i);//s
w[++cnt]=E(e[i][3],i);//e
}
sort(w+1,w+cnt+1,cmp);
deg[1]++;
for(i=1;i<=n;i++)tail[i]=-1,g[i].resize(deg[i]+2);
ins(1,0,0);//kx+b
for(i=1;i<=cnt;i++){
int x=w[i].y;
//printf("->%d %d\n",w[i].x,x);
if(x<0){
int y=-x;
dp[y]=cal(e[y][0],e[y][2]);
if(dp[y]<inf)dp[y]+=1LL*e[y][2]*e[y][2];
//printf("! %d %lld\n",y,dp[y]);
}else{
//printf("ins %d %lld\n",e[x][1],dp[x]);
ins(e[x][1],-2LL*e[x][3],dp[x]+1LL*e[x][3]*e[x][3]);
}
}
for(i=1;i<=m;i++)if(e[i][1]==n)ans=min(ans,dp[i]);
printf("%lld",ans);
}
/*
3 5
1 1 10 20
1 2 30 40
1 2 50 60
1 2 70 80
2 3 90 95
5 8
1 2 1 10
2 4 11 16
2 1 9 12
3 5 28 100
1 2 3 8
4 3 20 21
1 3 13 27
3 5 23 24
*/
I. Long Long Strings
初始串取字符集無窮的超長字串是最壞情況，壓縮區間後暴力操作即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
const LL inf = 1e12;
struct A
{
LL st;
LL ed;
LL stval;
bool operator < (const A &b)const
{
return st < b.st;
}
};
set<A>sot1, sot2;
void doit(set<A> &sot)
{
char ch;
while(~scanf(" %c", &ch))
{
if(ch == 'D')
{
LL pos;
scanf("%lld", &pos);
auto it = --sot.upper_bound({pos});
LL st = it->st;
LL ed = it->ed;
LL stval = it->stval;
sot.erase(it);
vector<A>vt;
if(st <= pos - 1)
{
vt.push_back({st, pos - 1, stval});
}
if(pos + 1 <= ed)
{
vt.push_back({pos, ed - 1, stval + pos + 1 - st});
}
for(auto w=sot.lower_bound({pos + 1}),nxt = w;w!= sot.end();w=nxt)
{
nxt = w; ++nxt;
vt.push_back({w->st - 1, w->ed - 1, w->stval});
sot.erase(w);
}
for(auto it : vt)sot.insert(it);
}
else if(ch == 'I')
{
char val_;
LL pos, val;
scanf("%lld %c", &pos, &val_);
val = inf + val_;
auto it = --sot.upper_bound({pos});
LL st = it->st;
LL ed = it->ed;
LL stval = it->stval;
sot.erase(it);
vector<A>vt;
if(st <= pos - 1)
{
vt.push_back({st, pos - 1, stval});
}
if(pos <= ed)
{
vt.push_back({pos + 1, ed + 1, stval + pos - st});
}
for(auto w=sot.lower_bound({pos + 1}),nxt = w;w!= sot.end();w=nxt)
{
nxt = w; ++nxt;
vt.push_back({w->st + 1, w->ed + 1, w->stval});
sot.erase(w);
}
vt.push_back({pos, pos, val});
for(auto it : vt)sot.insert(it);
}
else break;
/*
for(auto it : sot)
{
printf("%lld %lld %lld\n", it.st, it.ed, it.stval);
}
puts("show end");
*/
}
}
bool check()
{
LL ed1 = (--sot1.end())->ed;
LL ed2 = (--sot1.end())->ed;
if(ed1 != ed2)return 0;
LL now = 1;
while(now <= ed1)
{
auto it1 = sot1.begin();
auto it2 = sot2.begin();
LL ed1 = it1->ed;
LL ed2 = it2->ed;
if(it1->stval != it2->stval)return 0;
LL stval = it1->stval;
LL nxt = min(it1->ed, it2->ed) + 1;
sot1.erase(it1);
sot2.erase(it2);
if(nxt <= ed1)
{
sot1.insert({nxt, ed1, nxt - now + stval});
}
if(nxt <= ed2)
{
sot2.insert({nxt, ed2, nxt - now + stval});
}
now = nxt;
}
return 1;
}
int main()
{
sot1.clear();
sot1.insert({1, inf, 1});
sot2.clear();
sot2.insert({1, inf, 1});
doit(sot1);
doit(sot2);
/*
for(auto it : sot1)
{
printf("%lld %lld %lld\n", it.st, it.ed, it.stval);
}
puts("-------------");
for(auto it : sot2)
{
printf("%lld %lld %lld\n", it.st, it.ed, it.stval);
}
*/
puts(check() ? "0" : "1");
return 0;
}
/*
【trick&&吐槽】
【題意】
【分析】
【時間複雜度&&優化】
*/
J. Grid Coloring
設$f[i][j]$表示考慮前$i$行，第$i$行前$j$個都是藍色，後面都是紅色的方案數，暴力轉移即可。
時間複雜度$O(n^3)$。
#include<cstdio>
typedef long long ll;
const int N=40;
int n,m,i,j,k;
ll ans,f[N][N];
char a[N][N];
bool can[N][N];
inline bool check(int x,int y){
for(int i=1;i<=y;i++)if(a[x][i]=='R')return 0;
for(int i=y+1;i<=m;i++)if(a[x][i]=='B')return 0;
return 1;
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%s",a[i]+1);
for(i=1;i<=n;i++){
for(j=0;j<=m;j++){
can[i][j]=check(i,j);
}
}
//f[i][j] [1..j] is B, j+1..m is R
for(j=0;j<=m;j++)f[1][j]=can[1][j];
for(i=1;i<n;i++)for(j=0;j<=m;j++)for(k=0;k<=j;k++)f[i+1][k]+=f[i][j]*can[i+1][k];
for(j=0;j<=m;j++)ans+=f[n][j];
printf("%lld",ans);
}
K. Spinning Up Palindromes
顯然最優解中一定是從右往左操作，故問題可以轉化成：給定數字$A$，找到一個數位和最小的數字$B$，使得$A+B$是迴文串。
考慮從兩邊往中間DP，設$f[i][j][k]$表示考慮前後$i$個位置，前面希望後面進位為$j$，後面對前面進位為$k$的最小代價。
#include<cstdio>
#include<cstring>
const int N=50,inf=10000000;
int n,m,l,r,i,j,k,x,y,t,w;
char s[N];
int a[N],f[N][2][2],ans=inf;
inline void up(int&a,int b){a>b?(a=b):0;}
int main(){
scanf("%s",s+1);
n=strlen(s+1);
for(i=1;i<=n;i++)a[i]=s[i]-'0';
for(i=0;i<=n+5;i++)for(j=0;j<2;j++)for(k=0;k<2;k++)f[i][j][k]=inf;
f[0][0][0]=f[0][1][0]=0;
for(l=1,r=n;l<r;l++,r--)for(i=0;i<2;i++)for(j=0;j<2;j++)if(f[l-1][i][j]<inf){
w=f[l-1][i][j];
for(x=0;x<10;x++)for(y=0;y<10;y++){//what is b
for(k=0;k<2;k++){//jinwei from l+1
int A=a[l]+x+k,nA=0;
if(A>=10)A-=10,nA++;
if(nA!=i)continue;
int B=a[r]+y+j,nB=0;
if(B>=10)B-=10,nB++;
if(A!=B)continue;
up(f[l][k][nB],w+x+y);
}
}
}
m=n/2;
if(n%2){
for(i=0;i<2;i++)for(j=0;j<2;j++)if(f[m][i][j]<inf){
w=f[m][i][j];
for(x=0;x<10;x++){
if((a[m+1]+x+j)/10!=i)continue;
up(ans,w+x);
}
}
}else{
for(i=0;i<2;i++)for(j=0;j<2;j++)if(f[m][i][j]<inf){
w=f[m][i][j];
if(i==j)up(ans,w);
}
}
printf("%d",ans);
}
L. Delayed Work
暴力列舉人數即可。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll K,P,X,i;double ans;
int main(){
scanf("%lld%lld%lld",&K,&P,&X);
ans=1e20;
for(i=1;i<=3000000;i++){
ans=min(ans,1.0*K*P/i+1.0*X*i);
}
printf("%.3f",ans);
}
M. Unsatisfying
若初始2-SAT無解，那麼答案顯然為$0$。
如果不存在$\neg x\lor\neg y$，那麼無解。
否則列舉每個$x$，強行規定$x=true$，若2-SAT無解則答案為$1$。
否則答案只能為$2$。
時間複雜度$O(n(n+m))$。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 4010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m, nn;
int x, y;
vector<int> a[N];
bool pick[N];
int sta[N], top;
bool dfs(int x)
{
if(pick[x ^ 1]) return 0;
if(pick[x]) return 1;
pick[x] = 1;
sta[++ top] = x;
for(int i = a[x].size() - 1; ~ i; -- i){
int y = a[x][i];
if(!dfs(y)) return 0;
}
return 1;
}
bool solve()
{
for(int i = 0; i < nn; i += 2) if(!pick[i] && !pick[i ^ 1]){
top = 0;
if(!dfs(i)){
while(top) pick[sta[top --]] = 0;
top = 0;
if(!dfs(i ^ 1)) return 0;
}
}
return 1;
}
int main()
{
scanf("%d%d", &n, &m);
int flag = 0;
nn = n * 2;
for(int i = 0; i < nn; i ++) pick[i] = 0, a[i].clear();
for(int i = 1; i <= m; i ++){
scanf("%d%d", &x, &y);
int xx = abs(x), yy = abs(y);
xx *= 2; yy *= 2;
xx --; yy --;
if(x < 0 && y < 0) flag = 1;
if(x < 0 && y < 0){
a[xx].push_back(yy ^ 1);
a[yy].push_back(xx ^ 1);
//printf("%d %d\n", xx, yy ^ 1);
//printf("%d %d\n", yy, xx ^ 1);
}
else if(x < 0 && y > 0){
a[xx].push_back(yy);
a[yy ^ 1].push_back(xx ^ 1);
//printf("%d %d\n", xx, yy);
//printf("%d %d\n", yy ^ 1, xx ^ 1);
}
else if(x > 0 && y < 0){
a[yy].push_back(xx);
a[xx ^ 1].push_back(yy ^ 1);
//printf("%d %d\n", yy, xx);
//printf("%d %d\n", xx ^ 1, yy ^ 1);
}
else if(x > 0 && y > 0){
a[xx ^ 1].push_back(yy);
a[yy ^ 1].push_back(xx);
//printf("%d %d\n", xx ^ 1, yy);
//printf("%d %d\n", yy ^ 1, xx);
}
}
if(!flag){puts("-1"); return 0;}
if(!solve()) {puts("0"); return 0;}
for(int i = 1; i <= n; i ++){
for(int j = 0; j < nn; j ++){
//a[i].clear();
pick[j] = 0;
}
int xx = i * 2 - 1;
//a[xx].push_back(xx ^ 1);
a[xx ^ 1].push_back(xx);
if(!solve()){
puts("1"); return 0;
}
//a[xx].pop_back();
a[xx ^ 1].pop_back();
}
puts("2");
return 0;
}
/*
【trick&&吐槽】
【題意】
4 5
1 2
0 3
2 1
-1 -3
1 4
5 0
-2 3
3 5
4 2
3 -4
7 5
4 6
-2 -3
3 4
5 2
0
【分析】
【時間複雜度&&優化】
*/