XIII Open Grodno SU Championship
A. Alice in the Wonderland
按题意模拟。
#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;
struct A
{
int x, y, z;
}t, tt, st, ed;
queue<A> q;
int n, m, h;
char s[60][60][60];
bool e[60][60][60];
const int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
int main()
{
scanf("%d%d%d", &n, &m, &h);
for(int i = 1; i <= h; i ++){
for(int j = 1; j <= n; j ++){
scanf("%s", s[i][j] + 1);
}
}
for(int i = 1; i <= h; i ++){
for(int j = 1; j <= n; j ++){
for(int k = 1; k <= m; k ++){
if(s[i][j][k] == 'A'){
st.x = j; st.y = k; st.z = i;
}
else if(s[i][j][k] == 'E'){
ed.x = j; ed.y = k; ed.z = i;
}
}
}
}
int flag = 0;
q.push(st);
e[st.z][st.x][st.y] = 1;
while(! q.empty()){
t = q.front(); q.pop();
if(s[t.z][t.x][t.y] == 'E'){
flag = 1;
break;
}
if(s[t.z][t.x][t.y] == 'w'){
int i;
for(i = t.z; i <= h; i ++){ // 这里的方向要确认一下
if(s[i][t.x][t.y] != 'w'){
break;
}
} i --;
if(e[i][t.x][t.y] == 0){
tt.z = i; tt.x = t.x; tt.y = t.y;
q.push(tt);
e[i][t.x][t.y] = 1;
}
if(i != t.z) continue;
}
if(s[t.z][t.x][t.y] == 's'){
for(int i = t.z; i <= h; i ++){
if(s[i][t.x][t.y] == 's' && e[i][t.x][t.y] == 0){
e[i][t.x][t.y] = 1;
tt.z = i; tt.x = t.x; tt.y = t.y;
q.push(tt);
}
else if(s[i][t.x][t.y] != 's') break;
}
for(int i = t.z; i >= 1; i --){
if(s[i][t.x][t.y] == 's' && e[i][t.x][t.y] == 0){
e[i][t.x][t.y] = 1;
tt.z = i; tt.x = t.x; tt.y = t.y;
q.push(tt);
}
else if(s[i][t.x][t.y] != 's') break;
}
}
for(int i = 0; i < 4; i ++){
tt.x = t.x + dx[i];
tt.y = t.y + dy[i];
tt.z = t.z;
if(tt.x >= 1 && tt.x <= n && tt.y >= 1 && tt.y <= m && e[tt.z][tt.x][tt.y] == 0){
q.push(tt);
e[tt.z][tt.x][tt.y] = 1;
}
}
}
if(flag) puts("Yes"); else puts("No");
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
3 3 3
A..
.w.
...
...
.wE
...
...
.w.
...
【时间复杂度&&优化】
3 3 3
...
s.E
...
...
s..
...
...
s.A
...
*/
B. Batrachomyomachia
贪心，每次选承受能力最小的可行的。
#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 = 210, M = 1e4 + 10, 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, w;
const double eps = 1e-12;
int a[N][N], b[M];
double f[N][N];
bool del[111111];
multiset<int> sot;
multiset<int> :: iterator it, ir;
int sgn(double x){
if(fabs(x) < eps) return 0;
return x > 0 ? 1 : -1;
}
int main()
{
scanf("%d%d", &n, &w);
for(int i = 1; i < n; i ++) scanf("%d", &b[i]);
sort(b+1,b+n);
for(int i = 1; i <= 200; i ++){
for(int j = 1; j <= i; j ++){
a[i][j] = w;
}
}
for(int i = 1; i <= 200; i ++){
for(int j = 1; j <= i; j ++){
f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) / 2 + (a[i - 1][j] + a[i - 1][j - 1]) / 2;
}
}
for(int i=1;i<=200;i++){
sort(f[i] + 1, f[i] + i + 1);
reverse(f[i] + 1, f[i] + i + 1);
}
/*
for(int i = 1; i <= 10; i ++){
for(int j = 1; j <= i; j ++){
printf("%.0f ", f[i][j]);
}puts("");
}
*/
int ans = 0;
for(int i = 2; i <= 200; i ++){
for(int j = 1; j <= i; j ++){
int flag=0;
for(int k=1;k<n;k++)if(!del[k]&&sgn(b[k] - f[i][j])>=0){
flag=k;
break;
}
if(!flag){
ans = i - 1;
break;
}
del[flag]=1;
}
if(ans) break;
}
printf("%d\n", ans);
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
C. Cherries
将所有数字排序，那么一定是选取连续$B-A+1$个数进行配对，枚举所有方案即可。
#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 = 5050, 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 a[N];
int main()
{
while(~scanf("%d", &n))
{
int A, B;
scanf("%d%d", &A, &B);
for(int i = 1; i <= n; ++i)
{
scanf("%d", &a[i]);
}
sort(a + 1, a + n + 1);
int len = B - A;
LL ans = 1e18;
for(int i = 1; i + len <= n; ++i)
{
int j = i + len;
LL tmp = 0;
for(int k = i, x = A; k <= j; ++k, ++x)
{
tmp += abs(a[k] - x);
}
gmin(ans, tmp);
}
printf("%lld\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
D. Divisibility Game
预处理出约数集合后爆搜+卡时即可通过。
#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;
const int N=401;
const int lim=7;
const int lim2=3;
int i,j,x,now,ans,sum,ave,n,a[N],q[N];
vector<int>d[N];
int ED1 = CLOCKS_PER_SEC * 0.65;
int ED2 = CLOCKS_PER_SEC * 0.95;
void dfs(int s,int x){
if(x>n){
for(int i=1;i<=n;i++)printf("%d ",q[i]);
exit(0);
}
if(clock() > ED1)
{
return;
//puts("-1");
//exit(0);
}
for(vector<int>::iterator w=d[s].begin();w!=d[s].end();w++)
for(int j=min(a[*w],1);j;j--){
a[*w]-=j;
for(int o=0;o<j;o++)q[x+o]=*w;
dfs(s+(*w)*j,x+j);
a[*w]+=j;
}
}
void dfs2(int s,int x){
if(x>n){
for(int i=1;i<=n;i++)printf("%d ",q[i]);
exit(0);
}
if(clock() > ED2)
{
puts("-1");
exit(0);
}
for(vector<int>::iterator w=d[s].begin();w!=d[s].end();w++)
for(int j=min(a[*w],1);j;j--){
a[*w]-=j;
for(int o=0;o<j;o++)q[x+o]=*w;
dfs2(s+(*w)*j,x+j);
a[*w]+=j;
}
}
int main(){
scanf("%d",&n);
for(i=0;i<n;i++)scanf("%d",&x),a[x]++,sum+=x;
for(i=0;i<=sum;i++){
for(j=lim+1;j<=13;j++)
// for(j=13;j;j--)
if(i%j==0&&a[j])d[i].push_back(j);
for(j=lim2+1;j<=lim;j++)
// for(j=13;j;j--)
if(i%j==0&&a[j])d[i].push_back(j);
for(j=lim2;j;j--)
if(i%j==0&&a[j])d[i].push_back(j);
}
dfs(0,1);
for(i=0;i<=sum;i++)reverse(d[i].begin(),d[i].end());
dfs2(0,1);
puts("-1");
}
E. Enter the Word
设$dp[i]$表示打出前$i$个字符的最小代价，那么有$dp[i-1]\leq dp[i]\leq dp[i-1]+1$。
为了检查是否可以不$+1$，找到$dp$值的分界线，那么只要后面部分是前面部分的子串即可。
设$f[i]$表示子串匹配结束位置是$i$是否可行，可以通过bitset加速。
时间复杂度$O(\frac{n^2}{64})$。
#include<cstdio>
#include<bitset>
#include<cstring>
using namespace std;
const int N=200010;
int n,i,r,ans;char a[N];bitset<N>f,v[26];
int main(){
scanf("%s",a);
n=strlen(a);
for(i=0;i<n;i++){
a[i]-='a';
f=f<<1&v[a[i]];
if(!f.any()){
for(int k=r;k<i;k++)v[a[k]][k]=1;
r=i;
ans++;
f=v[a[i]];
}
}
printf("%d",ans);
}
F. Formula 1
按题意模拟即可，记录每个人的排名以及领先的圈数。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100010;
int n,m,i,x,f[N],pos[N],g[N];
bool cmp(int x,int y){return g[x]==g[y]?pos[x]<pos[y]:g[x]>g[y];}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++){
f[i]=i;
pos[i]=i;
}
while(m--){
scanf("%d",&x);
for(i=1;i<=n;i++)if(f[i]==x)break;
int o=i;
if(o==1){
g[x]++;
for(i=1;i<n;i++)f[i]=f[i+1];
f[n]=x;
swap(f[n],f[n-1]);
}else swap(f[o],f[o-1]);
//for(i=1;i<=n;i++)printf("%d ",f[i]);puts("");
}
for(i=1;i<=n;i++)pos[f[i]]=i;
sort(f+1,f+n+1,cmp);
for(i=1;i<=n&&i<=6;i++)printf("%d ",f[i]);
}
G. Game with Coins
将过程倒过来，则变成对于最后一个数，找到倒数第二个数，然后中间的数都要被最后一个数覆盖掉，区间DP即可。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=110;
int n,i,j,a[N],f[N][N],ans;
int dfs(int l,int r){
if(~f[l][r])return f[l][r];
int&t=f[l][r];
t=0;
int left=l+1,right=r;
if(left>right)right+=n;
for(int i=left;i<=right;i++)t=max(t,dfs(l,(i-1+n)%n)+dfs(i%n,r)+abs(a[l]-a[i%n]));
return t;
}
int main(){
scanf("%d",&n);
for(i=0;i<n;i++)scanf("%d",&a[i]);
for(i=0;i<n;i++)for(j=0;j<n;j++)if(i!=j)f[i][j]=-1;
for(i=0;i<n;i++)for(j=0;j<n;j++)ans=max(ans,dfs(i,j));
printf("%d",ans);
}
H. Hamnattan
首先特判起点终点都在同一条街道上的情况。其余情况Dijkstra求最短路即可，需要现算代价。
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
typedef pair<ll,P>PI;
typedef pair<ll,PI>PII;
const ll inf=1LL<<60;
const int N=110,M=500000;
int sa[N],sb[N];
int n,m,i,j,x,y,a[N],b[N],ns[N][N],ew[N][N],s[N][N];
int sx,sy,ex,ey;
ll d[N][N];
int g[N][N],v[M][2],w[M][2],nxt[M],ed;
priority_queue<PI,vector<PI>,greater<PI> >q;
ll ans=inf;
inline void add(int x,int y,int xx,int yy,int z,int zz){
v[++ed][0]=xx;
v[ed][1]=yy;
w[ed][0]=z;
w[ed][1]=zz;
nxt[ed]=g[x][y];
g[x][y]=ed;
}
inline void add2(int x,int y,int xx,int yy,int w,int ww){
add(x,y,xx,yy,w,ww);
add(xx,yy,x,y,w,ww);
}
inline int col(int x,int y,ll z){
int NS=ns[x][y],EW=ew[x][y],S=s[x][y];
z%=NS+EW;
if(S)return z<EW;
return z>=NS;
}
inline ll cal(int x,int y,int dir,ll z){
while(col(x,y,z)!=dir)z++;
return z;
}
inline void ext(int x,int y,ll z){
if(d[x][y]>z)q.push(PI(d[x][y]=z,P(x,y)));
}
void CHECK(){
int i,j;
for(i=1;i<=n;i++)for(j=1;j<=m;j++){
if(sy==ey)if(i<n)if(sb[j-1]==sy)if(sa[i-1]<=sx&&sx<=sa[i])
if(sa[i-1]<=ex&&ex<=sa[i]){
printf("%d",abs(sx-ex)+abs(sy-ey));
exit(0);
}
if(sx==ex)if(j<m)if(sa[i-1]==sx)if(sb[j-1]<=sy&&sy<=sb[j])
if(sb[j-1]<=ey&&ey<=sb[j]){
printf("%d",abs(sx-ex)+abs(sy-ey));
exit(0);
}
}
}
void EXT(int x,int y){
int i,j;
for(i=1;i<=n;i++)for(j=1;j<=m;j++){
if(i<n)if(sb[j-1]==y)if(sa[i-1]<=x&&x<=sa[i]){
ext(i,j,cal(i,j,1,x-sa[i-1]));
ext(i+1,j,cal(i+1,j,1,sa[i]-x));
}
if(j<m)if(sa[i-1]==x)if(sb[j-1]<=y&&y<=sb[j]){
ext(i,j,cal(i,j,0,y-sb[j-1]));
ext(i,j+1,cal(i,j+1,0,sb[j]-y));
}
}
}
void FIN(int x,int y){
int i,j;
for(i=1;i<=n;i++)for(j=1;j<=m;j++){
if(i<n)if(sb[j-1]==y)if(sa[i-1]<=x&&x<=sa[i]){
ans=min(ans,d[i][j]+x-sa[i-1]);
ans=min(ans,d[i+1][j]+sa[i]-x);
}
if(j<m)if(sa[i-1]==x)if(sb[j-1]<=y&&y<=sb[j]){
ans=min(ans,d[i][j]+y-sb[j-1]);
ans=min(ans,d[i][j+1]+sb[j]-y);
}
}
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<n;i++){
scanf("%d",&a[i]);
sa[i]=sa[i-1]+a[i];
}
for(i=1;i<m;i++){
scanf("%d",&b[i]);
sb[i]=sb[i-1]+b[i];
}
for(i=1;i<=n;i++)for(j=1;j<=m;j++){
if(i<n)add2(i,j,i+1,j,a[i],1);
if(j<m)add2(i,j,i,j+1,b[j],0);
}
for(j=1;j<=m;j++)for(i=1;i<=n;i++)scanf("%d%d%d",&ns[i][j],&ew[i][j],&s[i][j]);
scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
CHECK();
for(i=1;i<=n;i++)for(j=1;j<=m;j++)d[i][j]=inf;
EXT(sx,sy);
while(!q.empty()){
PI t=q.top();q.pop();
int x=t.second.first,y=t.second.second;
if(d[x][y]<t.first)continue;
//printf("%d %d %lld\n",x,y,t.first);
for(i=g[x][y];i;i=nxt[i]){
ext(v[i][0],v[i][1],cal(v[i][0],v[i][1],w[i][1],t.first+w[i][0]));
}
}
FIN(ex,ey);
printf("%lld",ans);
}
/*
4 3
10 10 10
10 10
1 99 0
99 1 0
50 99 0
1 99 1
1 99 0
99 1 0
20 41 1
1 99 0
99 1 0
1 99 1
99 1 0
99 1 0
1 10
30 19
*/
I. Integer Pairs
只要$a[j]<0$即合法。
#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;
int main()
{
while(~scanf("%d", &n))
{
int neg = 0;
for(int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
neg += x < 0;
}
LL ans = neg * (n - 1ll);
printf("%lld\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
3
-1 -2 -3
*/
J. Jedi Training
线段树维护$f[l][r]$表示对应区间内选择子序列的左端点奇偶性为$l$，右端点奇偶性为$r$时的最大和。
#include<cstdio>
#include<algorithm>
using namespace std;
#define rep(i) for(int i=0;i<2;i++)
typedef long long ll;
const int N=100010,M=262150;
const ll inf=1LL<<60;
int n,m,i,a[N],op,x,y;
inline void up(ll&a,ll b){a<b?(a=b):0;}
struct E{
ll f[2][2];
E(){rep(i)rep(j)f[i][j]=-inf;}
void clr(){rep(i)rep(j)f[i][j]=-inf;}
void set(int x,ll p){
clr();
f[x&1][x&1]=p;
}
E operator+(const E&b){
E c;
rep(i)rep(j)c.f[i][j]=max(f[i][j],b.f[i][j]);
rep(i)rep(j)rep(k)up(c.f[i][k],f[i][j]+b.f[j^1][k]);
return c;
}
void write(){
ll t=-inf;
rep(i)rep(j)up(t,f[i][j]);
printf("%lld\n",t);
}
}v[M];
void build(int x,int a,int b){
if(a==b){
v[x].set(a,::a[a]);
return;
}
int mid=(a+b)>>1;
build(x<<1,a,mid),build(x<<1|1,mid+1,b);
v[x]=v[x<<1]+v[x<<1|1];
}
void change(int x,int a,int b,int c,int d){
if(a==b){
v[x].set(a,d);
return;
}
int mid=(a+b)>>1;
if(c<=mid)change(x<<1,a,mid,c,d);else change(x<<1|1,mid+1,b,c,d);
v[x]=v[x<<1]+v[x<<1|1];
}
E ask(int x,int a,int b,int c,int d){
if(c<=a&&b<=d)return v[x];
int mid=(a+b)>>1;
E t;
t.clr();
if(c<=mid)t=ask(x<<1,a,mid,c,d);
if(d>mid)t=t+ask(x<<1|1,mid+1,b,c,d);
return t;
}
int main(){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%d",&a[i]);
build(1,1,n);
while(m--){
scanf("%d%d%d",&op,&x,&y);
if(op==1)change(1,1,n,x,y);
else{
ask(1,1,n,x,y).write();
}
}
}
K. Kings of a Round Table
假设$1$号国王一定位于$1$号位置，并不区分剩下$7$个国王，则这种情况下方案数还要乘以$n\times 7!$。
那么剩下的方案数大约只有$3\times 10^8$个，爆搜打表即可。
#include<cstdio>
long long f[111];
int n;
int main(){
f[9]=0;
f[10]=0;
f[11]=0;
f[12]=0;
f[13]=0;
f[14]=0;
f[15]=0;
f[16]=0;
f[17]=685440;
f[18]=725760;
f[19]=11491200;
f[20]=6451200;
f[21]=83825280;
f[22]=38142720;
f[23]=397837440;
f[24]=170311680;
f[25]=1441440000;
f[26]=617460480;
f[27]=4330609920;
f[28]=1905684480;
f[29]=11330323200;
f[30]=5175878400;
f[31]=26645794560;
f[32]=12675317760;
f[33]=57564017280;
f[34]=28504707840;
f[35]=116035920000;
f[36]=59698114560;
f[37]=220799779200;
f[38]=117723513600;
f[39]=400156848000;
f[40]=220502016000;
f[41]=695520483840;
f[42]=395054150400;
f[43]=1165870379520;
f[44]=680891904000;
f[45]=1893253824000;
f[46]=1134285546240;
f[47]=2989486241280;
f[48]=1833544581120;
f[49]=4604213577600;
f[50]=2885462496000;
f[51]=6934509429120;
f[52]=4433085296640;
f[53]=10236190124160;
f[54]=6664974140160;
f[55]=14837041296000;
f[56]=9826142699520;
f[57]=21152159804160;
f[58]=14230860215040;
f[59]=29701625184000;
f[60]=20277521510400;
f[61]=41130725126400;
f[62]=28465795572480;
f[63]=56232969811200;
f[64]=39416274616320;
f[65]=75976140240000;
f[66]=53892855878400;
f[67]=101531626035840;
f[68]=72828098703360;
f[69]=134307318499200;
f[70]=97351809811200;
scanf("%d",&n);
printf("%lld",f[n]);
}
L. Lines and Polygon
求出直线与凸包的交点，然后在附近枚举即可得到最近点。
#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 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 = 1e5 + 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;
const long double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
return x > 0 ? 1 : -1;
}
struct point
{
long double x, y;
point(){}
point(long double a, long double b){x = a; y = b;}
long double det (const point &b)const{
return x * b.y - y * b.x;
}
friend bool operator < (const point &a, const point &b){
if(sgn(a.x - b.x) == 0) return sgn(a.y - b.y) < 0;
return sgn(a.x - b.x) < 0;
}
friend point operator - (const point &a, const point &b){
return point(a.x - b.x, a.y - b.y);
}
};
struct Convex
{
int n;
vector<point> a, upper, lower;
Convex(){}
Convex (vector<point> _a) : a(_a){
n = a.size();
int ptr = 0;
for(int i = 1; i < n; i ++) if(a[ptr] < a[i]) ptr = i;
for(int i = 0; i <= ptr; i ++) lower.push_back(a[i]);
for(int i = ptr; i < n; i ++) upper.push_back(a[i]);
upper.push_back(a[0]);
}
int sign(long double x){
if(fabs(x) < eps) return 0;
return x > 0 ? 1 : -1;
}
pair<long double, int> get_tangent(vector<point> &convex, point vec){
int l = 0, r = (int) convex.size() - 2;
for(; l + 1 < r;){
int mid = (l + r) / 2;
if(sign((convex[mid + 1] - convex[mid]).det(vec)) > 0) r = mid;
else l = mid;
}
return max(make_pair(vec.det(convex[r]), r), make_pair(vec.det(convex[0]), 0));
}
int binary_search(point u, point v, int l, int r){
int sl = sign((v - u).det(a[l % n] - u));
for(; l + 1 < r;){
int mid = (l + r) / 2;
int smid = sign((v - u).det(a[mid % n] - u));
if(smid == sl) l = mid;
else r = mid;
}
return l % n;
}
int get_tangent(point vec){
pair<long double, int> ret = get_tangent(upper, vec);
ret.second = (ret.second + (int)lower.size() - 1) % n;
ret = max(ret, get_tangent(lower, vec));
return ret.second;
}
bool get_intersection(point u, point v, int &i0, int &i1){
int p0 = get_tangent(u - v), p1 = get_tangent(v - u);
if(sign((v - u).det(a[p0] - u)) * sign((v - u).det(a[p1] - u)) < 0){
if(p0 > p1) swap(p0, p1);
i0 = binary_search(u, v, p0, p1);
i1 = binary_search(u, v, p1, p0 + n);
return true;
}
else{
return false;
}
}
};
int n;
point p[N];
vector<point> a, b;
Convex D;
int m;
long double A, B, C;
long double cal(int i0)
{
return fabs((A * D.a[i0].x + B * D.a[i0].y + C) );
}
const double INF = 1e9;
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i ++) {
//scanf("%lf%lf", &p[i].x, &p[i].y);
double x, y;
scanf("%lf%lf", &x, &y);
p[i].x = x; p[i].y = y;
//a.push_back(p[i]);
}
for(int i = n - 1; i >= 0; i --) a.push_back(p[i]);
int ptr = 0;
for(int i = 1; i < n; i ++){
if(a[ptr] < a[i]) ptr = i;
}
for(int i = ptr; i < n; i ++){
b.push_back(a[i]);
}
for(int i = 0; i < ptr; i ++){
b.push_back(a[i]);
}
D = Convex(b);
scanf("%d", &m);
for(int i = 1; i <= m; i ++){
double AA, BB, CC;
scanf("%lf%lf%lf", &AA, &BB, &CC);
A = AA; B = BB; C = CC;
double ans = 1e18;
int i0, i1;
point p0, p1;
if(A){
p0.y = 0, p0.x = -C / A;
p1.y = INF, p1.x = (- C - B * INF) / A;
}
else if(B){
p0.x = 0, p0.y = -C / B;
p1.x = INF, p1.y = (-C - INF * A) / B;
}
//else while(1);
if(D.get_intersection(p0, p1, i0, i1)){
#define next(i) ((i + 1) % n)
#define pre(i) ((i - 1 + n) % n)
for(int j = 0; j < 10; j ++){
gmin(ans, cal(i0));
i0 = next(i0);
}
for(int j = 0; j < 20; j ++){
gmin(ans, cal(i0));
i0 = pre(i0);
}
for(int j = 0; j < 10; j ++){
gmin(ans, cal(i1));
i1 = next(i1);
}
for(int j = 0; j < 20; j ++){
gmin(ans, cal(i1));
i1 = pre(i1);
}
}
else{
//ans = 0;
//while(1);
for(int j = 0; j < n; j ++) gmin(ans, cal(j));
}
double ANS = ans / sqrt(A * A + B * B);
printf("%.6f\n", ANS);
}
return 0;
}
/*
【trick&&吐槽】
4
1 3
3 1
1 -1
-1 1
1
0 4 -5
【题意】
【分析】
【时间复杂度&&优化】
*/
M. MIPT Campus
留坑。
