A. Airport Logistics
根据光路最快原理以及斯涅尔定律，可以得到从定点$P$进入某条直线的最佳入射角。
求出每个端点到每条线段的最佳点，建图求最短路即可。
时间复杂度$O(n^2log n)$。
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef pair<double,int>PI;
const int N=41000,M=1000000;
const double eps=1e-9,inf=1e100;
int n,cnt,i,j,g[N],v[M],nxt[M],ed;double w[M],d[N],va,vb,si[2],co[2];
priority_queue<PI,vector<PI>,greater<PI> >q;
inline void add(int x,int y,double z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
inline void add2(int x,int y,double z){add(x,y,z),add(y,x,z);}
inline void ext(int x,double y){if(y+eps<d[x])q.push(PI(d[x]=y,x));}
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);}
double operator*(P b){return x*b.x+y*b.y;}
bool operator==(P b){return !sgn(x-b.x)&&!sgn(y-b.y);}
double len(){return hypot(x,y);}
P rotate(double s,double c){return P(x*c-y*s,x*s+y*c);}
P rot90(){return P(-y,x);}
}a[210];
struct E{
double x;int y;
E(){}
E(double _x,int _y){x=_x,y=_y;}
}e[410];
inline bool cmp(const E&a,const E&b){return a.x<b.x;}
inline double cross(P a,P b){return a.x*b.y-a.y*b.x;}
inline bool point_on_segment(P p,P a,P b){
return sgn(cross(b-a,p-a))==0&&sgn((p-a)*(p-b))<=0;
}
inline P line_intersection(P a,P b,P p,P q){
double U=cross(p-a,q-p),D=cross(b-a,q-p);
return a+(b-a)*(U/D);
}
inline void work(int st,int en){
int i,j,m=2;
P A=a[st],B=a[en],C=(B-A).rot90();
e[1]=E(0,st),e[2]=E(C.len(),en);
for(i=0;i<=(n<<1|1);i++)if(i!=st&&i!=en)for(j=0;j<2;j++){
P D=line_intersection(A,B,a[i],a[i]+C.rotate(si[j],co[j]));
if(D==A||D==B)continue;
if(!point_on_segment(D,A,B))continue;
cnt++;
add2(i,cnt,(a[i]-D).len()/vb);
e[++m]=E((D-A).len(),cnt);
}
sort(e+1,e+m+1,cmp);
for(i=1;i<m;i++)add(e[i].y,e[i+1].y,(e[i+1].x-e[i].x)/va);
}
int main(){
scanf("%lf%lf",&a[0].x,&a[0].y);
scanf("%lf%lf",&a[1].x,&a[1].y);
scanf("%d",&n);
va=2,vb=1;
for(i=1;i<=n;i++){
scanf("%lf%lf",&a[i<<1].x,&a[i<<1].y);
scanf("%lf%lf",&a[i<<1|1].x,&a[i<<1|1].y);
}
si[0]=vb/va;
co[0]=sqrt(1.0-si[0]*si[0]);
si[1]=-si[0];
co[1]=co[0];
cnt=n<<1|1;
for(i=0;i<=cnt;i++)for(j=0;j<i;j++)add2(i,j,(a[i]-a[j]).len()/vb);
for(i=1;i<=n;i++)work(i<<1,i<<1|1);
for(i=0;i<=cnt;i++)d[i]=inf;
ext(0,0);
while(!q.empty()){
PI t=q.top();q.pop();
if(t.first-eps>d[t.second])continue;
for(i=g[t.second];i;i=nxt[i])ext(v[i],t.first+w[i]);
}
printf("%.10f",d[1]);
}
B. Battle Simulation
按题意模拟即可。
#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
const int top = 1e6;
const int N = top + 10, M = 262150;
char s[N], ans[N];
int main(){
scanf("%s", s);
int len = strlen(s);
int num = 0;
for(int i = 0; i < len; i ++){
if(i + 2 < len && s[i] != s[i + 1] && s[i] != s[i + 2] && s[i + 1] != s[i + 2]){
ans[++ num] = 'C';
i += 2;
}
else if(s[i] == 'R') ans[++ num] = 'S';
else if(s[i] == 'B') ans[++ num] = 'K';
else ans[++ num] = 'H';
}
printf("%s
", ans + 1);
}
/*
4
4 3
1 2
2 3
3 4
5
2 1 4
1 2 3
2 1 4
2 2 3
2 2 4
8 9
1 2
2 3
1 3
3 4
4 5
4 6
5 7
5 8
7 8
5
2 7 8
2 1 6
2 4 7
1 6 8
2 5 6
*/
C. Brexit
拓扑排序，不断删掉不合法的点即可。
#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() { 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 = 3e5 + 10, 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, ME, ST;
vector<int>a[N];
bool stay[N];
int ind[N];
int least[N];
int sta[N];
int main()
{
while(~scanf("%d%d%d%d", &n, &m, &ME, &ST))
{
for(int i = 1; i <= n; ++i)
{
a[i].clear();
stay[i] = 1;
ind[i] = 0;
}
for(int i = 1; i <= m; ++i)
{
int x, y; scanf("%d%d", &x, &y);
a[x].push_back(y);
a[y].push_back(x);
++ind[x];
++ind[y];
}
for(int i = 1; i <= n; ++i)
{
least[i] = ind[i] / 2 + 1;
}
int top = 0;
sta[++top] = ST;
stay[ST] = 0;
while(top)
{
int x = sta[top--];
for(auto y : a[x]) if(stay[y])
{
if(--ind[y] < least[y])
{
sta[++top] = y;
stay[y] = 0;
}
}
}
puts(stay[ME] ? "stay" : "leave");
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
4 3 4 1
2 3
2 4
1 2
5 5 1 1
3 4
1 2
2 3
1 3
2 5
4 5 3 1
1 2
1 3
2 3
2 4
3 4
10 14 1 10
1 2
1 3
1 4
2 5
3 5
4 5
5 6
5 7
5 8
5 9
6 10
7 10
8 10
9 10
*/
D. Bridge Automation
设$f[i]$表示前$i$艘船开走的最小代价，枚举与$i$最早一起开走的船$j$转移即可。
时间复杂度$O(n^2)$。
#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 = 4040, 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;
int t[N];
int f[N];
int main()
{
while(~scanf("%d", &n))
{
for(int i = 1; i <= n; ++i)
{
scanf("%d", &t[i]);
}
MS(f, 63); f[1] = 0;
for(int i = 1; i <= n; ++i)
{
//printf("%d
", f[i]);
int down = t[i] + 1800 - 60; //桥放下的时间是定的
int pre = t[i] + 1800 - 20;
for(int j = i; j <= n; ++j) //[i, j]一起走
{
pre = max(pre + 20, t[j]); //这艘船开始开走的最早时间
gmin(f[j + 1], f[i] + pre + 20 + 60 - down);
}
}
printf("%d
", f[n + 1]);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
2
100
200
3
100
200
2010
3
100
200
2100
*/
E. Charles in Charge
二分答案，最短路检验。
时间复杂度$O(nlog^2n)$。
#include<cstdio>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll,int>P;
const int N=10010,M=200010;
const ll inf=1LL<<50;
int n,m,rate,i,x,y,z,g[N],v[M],w[M],nxt[M],ed,l,r,mid,ans;
ll mindis,d[N];
inline void add(int x,int y,int z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
ll dij(int lim){
int i;
for(i=1;i<=n;i++)d[i]=inf;
priority_queue<P,vector<P>,greater<P> >q;
q.push(P(d[1]=0,1));
while(!q.empty()){
P t=q.top();q.pop();
if(d[t.second]>t.first)continue;
for(i=g[t.second];i;i=nxt[i])if(w[i]<=lim&&d[v[i]]>t.first+w[i])
q.push(P(d[v[i]]=t.first+w[i],v[i]));
}
return d[n];
}
int main(){
scanf("%d%d%d",&n,&m,&rate);
rate+=100;
for(i=1;i<=m;i++){
scanf("%d%d%d",&x,&y,&z);
add(x,y,z),add(y,x,z);
}
mindis=dij(1000000000);
mindis*=rate;
//should / 100
l=1,r=1000000000;
while(l<=r){
mid=(l+r)>>1;
if(dij(mid)*100<=mindis)r=(ans=mid)-1;else l=mid+1;
}
printf("%d",ans);
}
F. Endless Turning
模拟$30000$步，这其中必然存在循环节，找出循环节后最后零碎部分继续模拟即可。
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-9;
const int N=110;
const int inf=100000000;
long long T;
int n,m,i,pos,nxtpos;
char name[N][N];
int loop=inf;
inline int sgn(double x){
if(x<-eps)return -1;
if(x>eps)return 1;
return 0;
}
struct P{
double x,y;
P(){x=y=0;}
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);}
bool operator==(P b){return !sgn(x-b.x)&&!sgn(y-b.y);}
bool operator!=(P b){return sgn(x-b.x)||sgn(y-b.y);}
bool operator<(P b){
if(sgn(x-b.x))return x<b.x;
return y<b.y;
}
double operator*(P v){return x*v.x+y*v.y;}
double len(){return hypot(x,y);}
double len_sqr(){return x*x+y*y;}
}A,B,C,D,a[N],b[N];
struct E{
P x;
int y,z;
E(){}
E(P _x,int _y,int _z){x=_x,y=_y,z=_z;}
}e[100010];
inline bool cmp(E a,E b){
if(a.x!=b.x)return a.x<b.x;
return a.y!=b.y?a.y<b.y:a.z<b.z;
}
inline double cross(P a,P b){return a.x*b.y-a.y*b.x;}
inline bool point_on_segment(P p,P a,P b){
return sgn(cross(b-a,p-a))==0;
}
inline int line_intersection(P a,P b,P p,P q,P&o,double&t){
double U=cross(p-a,q-p);
double D=cross(b-a,q-p);
if(sgn(D)==0)return 0;
o=a+(b-a)*(U/D);
t=U/D;
return 1;
}
int getpos(P o){
for(int i=1;i<=n;i++)if(point_on_segment(o,a[i],b[i]))return i;
return 0;
}
inline bool go(){
int nxt=0;double dis=1e100;
for(int i=1;i<=n;i++)if(i!=abs(pos)){
P o;double t;
if(line_intersection(A,A+B,a[i],b[i],o,t)){
//printf("->%d %.8f %.8f %.8f
",i,o.x,o.y,t);
if(t<eps)continue;
if(t<dis)dis=t,nxt=i;
}
}
if(!nxt)return 0;
line_intersection(A,A+B,a[nxt],b[nxt],C,dis);
nxtpos=nxt;
D=b[nxt]-a[nxt];
if(cross(C-A,C+D-A)>=0)nxtpos*=-1,D=D*(-1);
return 1;
}
int main(){
scanf("%d%lld%lf%lf",&n,&T,&A.x,&A.y);
for(i=1;i<=n;i++){
scanf("%s",name[i]);
scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&b[i].x,&b[i].y);
}
pos=getpos(A);
if(T==0){
puts(name[abs(pos)]);
return 0;
}
B=b[pos]-a[pos];//direction
if(B.x<0)B=B*(-1),pos*=-1;
//printf("st : (%.8f,%.8f)->(%.8f,%.8f)
",A.x,A.y,A.x+B.x,A.y+B.y);
while(T>0&&m<30000){
if(!go()){
puts(name[abs(pos)]);
return 0;
}
A=C,B=D,pos=nxtpos;
//printf("(%.8f,%.8f)->(%.8f,%.8f)
",A.x,A.y,A.x+B.x,A.y+B.y);
T--;
m++;
e[m]=E(A,pos,m);
}
if(T==0){
puts(name[abs(pos)]);
return 0;
}
sort(e+1,e+m+1,cmp);
for(i=1;i<m;i++)if(e[i].x==e[i+1].x&&e[i].y==e[i+1].y){
loop=min(loop,e[i+1].z-e[i].z);
}
T%=loop;
//printf("loop=%d
",loop);
while(T>0){
go();
A=C,B=D,pos=nxtpos;
//printf("(%.8f,%.8f)->(%.8f,%.8f)
",A.x,A.y,A.x+B.x,A.y+B.y);
T--;
}
puts(name[abs(pos)]);
}
/*
3 4 1 1
Broadway 0 0 0 1
Narrowlane 0 0 1 0
Homedrive 1 1 2 0
3 4 5 0
Broadway 0 0 0 1
Narrowlane 0 0 1 0
Homedrive 1 1 2 0
*/
G. Manhattan Positioning System
将曼哈顿距离转切比雪夫距离，则可行解位于若干个正方形的交集上。
在对应正方形四个端点附近枚举所有点检查即可。
时间复杂度$O(n)$。
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const ll inf=1LL<<60;
int n,i;
ll a[1010][3],xl,xr,yl,yr;
ll retx,rety;
set<P>ans;
inline void check(ll X,ll Y){
ll x=(X+Y)/2,y=X-x;
for(int i=1;i<=n;i++)if(abs(x-a[i][0])+abs(y-a[i][1])!=a[i][2])return;
retx=x,rety=y;
ans.insert(P(x,y));
}
void go(ll x,ll y){
ll k=10;
for(ll i=x-k;i<=x+k;i++)for(ll j=y-k;j<=y+k;j++)check(i,j);
}
int main(){
scanf("%d",&n);
xl=-inf;
xr=inf;
yl=-inf;
yr=inf;
for(i=1;i<=n;i++){
ll x,y,d;
scanf("%lld%lld%lld",&x,&y,&d);
a[i][0]=x;
a[i][1]=y;
a[i][2]=d;
ll X=x+y,Y=x-y;
xl=max(xl,X-d);
xr=min(xr,X+d);
yl=max(yl,Y-d);
yr=min(yr,Y+d);
}
go(xl,yl);
go(xl,yr);
go(xr,yl);
go(xr,yr);
if(ans.size()==0)puts("impossible");
else if(ans.size()==1)printf("%lld %lld",retx,rety);
else puts("uncertain");
}
H. Multiplying Digits
最优解中从低位到高位每个数必定是不上升的，爆搜剪枝配合卡时即可通过。
#include<cstdio>
#include<time.h>
typedef unsigned long long ll;
const int N = 11111;
const ll inf = 1ULL << 63;
ll K, n, i, a[N], ans = inf;
int m;
inline ll mul(ll a, ll b) {
if (a>ans / b)return ans;
return a*b;
}
int ED = 5.9 * CLOCKS_PER_SEC;
inline bool cal(ll ret, ll x, ll base, ll now)
{
while (ret >= x)
{
now += mul(base, x);
base = mul(base, K);
if (now >= ans)return 0;
ret /= x;
}
return 1;
}
void dfs(int x, ll base, ll ret, ll n) {//now consider a[x]
//printf("%d %llu %llu %llu
",x,base,ret,n);
if (ret >= ans)return;
if (clock() > ED)return;
if (n == 1) {
//printf("%llu
",ret);
ans = ret;
return;
}
if (base >= ans)return;
while (1)
{
while (x <= m&&n%a[x])x++;
if (x>m)return;
ll t = ret + mul(base, a[x]);
if (t >= ans)return;
if (!cal(n, a[x], base, ret))return;
dfs(x, mul(base, K), t, n / a[x]);
if (clock() > ED)return;
x++;
}
}
int main() {
/* ll t=1;
for(i=1;i<=18;i++)t*=6;
printf("%llu
",t);*/
scanf("%llu%llu", &K, &n);
if (n == 1) {
puts("1");
return 0;
}
for (i = K - 1; i >= 2; i--)if (n%i == 0) {
a[++m] = i;
}
//for(i=1;i<=m;i++)printf("%llu ",a[i]);puts("");
//printf("%d
",m);
a[0] = a[m + 1] = 1;
dfs(1, 1, 0, n);
if (ans == inf)puts("impossible");
else printf("%llu", ans);
return 0;
}
/*
10 24
10 11
9 216
10000 5810859769934419200
9 101559956668416
*/
I. Older Brother
按题意模拟即可。
#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() { 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 = 0, 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;
bool solve()
{
if(n == 1)return 0;
for(int i = 2; i * i <= n; ++i)if(n % i == 0)
{
int x = n;
while(x % i == 0)
{
x /= i;
}
return x == 1;
}
return 1;
}
int main()
{
while(~scanf("%d", &n))
{
puts(solve() ? "yes" : "no");
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
J. Programming Tutors
二分答案，二分图匹配检验。
时间复杂度$O(n^3log n)$。
#include<cstdio>
const int N=110;
int n,i,j,l,r,mid,ans,d[N][N],v[N],f[N];
struct P{int x,y;}a[N],b[N];
inline int abs(int x){return x>0?x:-x;}
bool find(int x){
for(int i=1;i<=n;i++)if(d[x][i]<=mid&&!v[i]){
v[i]=1;
if(!f[i]||find(f[i]))return f[i]=x,1;
}
return 0;
}
bool check(){
int i,j;
for(j=1;j<=n;j++)f[j]=0;
for(i=1;i<=n;i++){
for(j=1;j<=n;j++)v[j]=0;
if(!find(i))return 0;
}
return 1;
}
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);
for(i=1;i<=n;i++)scanf("%d%d",&b[i].x,&b[i].y);
for(i=1;i<=n;i++)for(j=1;j<=n;j++)d[i][j]=abs(a[i].x-b[j].x)+abs(a[i].y-b[j].y);
l=0,r=1000000000;
while(l<=r){
mid=(l+r)>>1;
if(check())r=(ans=mid)-1;else l=mid+1;
}
printf("%d",ans);
}
K. Safe Racing
设$f[i]$表示长度为$i$的序列，$1$和$i$必选时的合法方案数，可以通过前缀和$O(n)$求出。
枚举第一个放的位置，那么最后一个位置的范围是一个区间，同样可以前缀和加速。
时间复杂度$O(n)$。
#include<cstdio>
const int N=1000010,P=123456789;
int n,m,i,j,ans,f[N],s[N];
int main(){
scanf("%d%d",&n,&m);
f[1]=s[1]=1;
for(i=2;i<=n;i++){
f[i]=s[i-1];
if(i-m-1>=0)f[i]-=s[i-m-1];
f[i]=(f[i]+P)%P;
s[i]=(s[i-1]+f[i])%P;
}
for(i=1;i<=m;i++){
j=n-m+i;
if(j>n)continue;
int l=j-i+1,r=n-i+1;
ans=(ans+s[r])%P;
ans=(ans-s[l-1]+P)%P;
}
printf("%d",ans);
}
L. Sticky Situation
排序后检查相邻$3$项能否形成三角形即可。
#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() { 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 = 2e5 + 10, 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;
LL a[N];
bool solve()
{
for(int i = 1; i <= n - 2; ++i)
{
if(a[i] + a[i + 1] > a[i + 2])return 1;
}
return 0;
}
int main()
{
while(~scanf("%d", &n))
{
for(int i = 1; i <= n; ++i)
{
scanf("%lld", &a[i]);
}
sort(a + 1, a + n + 1);
puts(solve() ? "possible" : "impossible");
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/