2021 ICPC 网络赛2
Posted 行码棋
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了2021 ICPC 网络赛2相关的知识，希望对你有一定的参考价值。
A
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e3 + 7;
int a[N], b[N], id[N];
int n, m, tot;
vector<int> G[N], ans;
int c[N];
int lowbit(int x) {
return x & -x;
}
void add(int x, int v) {
while (x <= n) {
c[x] += v;
x += lowbit(x);
}
}
int ask(int x) {
int res = 0;
while (x > 0) {
res += c[x];
x -= lowbit(x);
}
return res;
}
bool check() {
for (int i = 2; i <= n; ++i) {
if (a[i] < a[i - 1]) return false;
}
return true;
}
int casT;
void run() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
b[i] = a[i];
}
if (is_sorted(a + 1, a + 1 + n)) {
printf("0");
if (casT != 0) puts("");
return;
}
if (m == 1) {
printf("-2");
if (casT != 0) puts("");
return;
}
if (m == 2) {
int pre = 1;
while (pre < n && a[pre] <= a[pre + 1]) pre++;
int suf = pre + 1;
while (suf < n && a[suf] <= a[suf + 1]) suf++;
// printf("%d %d\\n", pre, suf);
if (suf != n || a[n] > a[1]) {
printf("-2");
if (casT != 0) puts("");
return;
} else {
puts("1\\n2");
printf("0 %d %d\\n", pre, n);
printf("2 1");
if (casT != 0) puts("");
return;
}
}
sort(b + 1, b + 1 + n);
tot = unique(b + 1, b + 1 + n) - b - 1;
ans.clear();
for(int i = 1; i <= n; ++i) {
c[i] = 0;
G[i].clear();
}
for (int i = 1; i <= n; ++i) {
id[i] = lower_bound(b + 1, b + 1 + tot, a[i]) - b;
G[ id[i] ].push_back(i);
}
// printf("%d\\n", n);
for (int i = tot; i >= 1; --i) {
for (auto v : G[i]) {
int pos = v;
pos += ask(pos);
if (pos == 1) continue;
// printf("%d\\n", pos);
assert(pos >= 1 && pos <= n);
ans.push_back(pos);
add(1, 1);
add(v, -1);
}
}
printf("%d\\n", ans.size());
int lst = ans.size();
for (auto v : ans) {
lst--;
if (v == n) {
puts("2");
printf("0 %d %d\\n", v - 1, v);
printf("2 1");
if (lst != 0 || casT != 0) puts("");
} else {
puts("3");
printf("0 %d %d %d\\n", v - 1, v, n);
printf("2 1 3");
if (lst != 0 || casT != 0) puts("");
}
}
}
int main() {
scanf("%d", &casT);
while (casT--) run();
return 0;
}
E
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double db;
const int N = 1e2+5;
const db pi = acos(-1);
const db eps = 1e-9;
struct P{
db x,y;
void read(){
cin>>x>>y;
}
friend P operator - (P a,P b){
return {a.x-b.x,a.y-b.y};
}
friend P operator + (P a,P b){
return {(a.x+b.x)/2,(a.y+b.y)/2};
}
db deg(){
db k=atan2(y,x);
if(k<0)k+=2*pi;
return k;
}
P rotate(double k){
db x1=x,y1=y;
double c=cos(k),s=sin(k);
x=x1*c-y1*s;
y=x1*s+y1*c;
}
}a[N],b[N];
int n;
vector<db> v;
db ans[N][N];
int main(){
cin>>n;
for(int i=1;i<=n;++i)a[i].read(),b[i]=a[i];
for(int c=1;c<=n;++c){
v.clear();
v.push_back(0);
v.push_back(2*pi);
for(int i=1;i<=n;++i)if(i!=c)
for(int j=1;j<i;++j)if(j!=c){
P t=a[i]-a[j];
db w;
w=pi/2-t.deg();
if(w<0)w+=2*pi;
v.push_back(w);
w=pi*3/2-t.deg();
if(w<0)w+=2*pi;
v.push_back(w);
t=a[i]+a[j];
t=t-a[c];
if(abs(t.x)<eps&&abs(t.y)<eps)continue;
w=pi/2-t.deg();
if(w<0)w+=2*pi;
v.push_back(w);
w=pi*3/2-t.deg();
if(w<0)w+=2*pi;
v.push_back(w);
}
sort(v.begin(),v.end());
for(int j=1;j<=n;++j)a[j]=b[j];
for(int i=1;i<v.size();++i){
db det=(v[i]-v[i-1])/2;
for(int j=1;j<=n;++j)a[j].rotate(det);
int p=0;
db res=0;
for(int k=1;k<=n;++k)
if(k!=c){
P t=a[c]-a[k];
if(!p||abs(t.x)<res-eps){
p=k;
res=abs(t.x);
}
}
ans[c][p]+=det*2;
for(int j=1;j<=n;++j)a[j].rotate(det);
}
}
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j)printf("%.11Lf ",ans[i][j]/2/pi);
if(i!=n)printf("\\n");
}
}
J
#include<bits/stdc++.h>
using namespace std;
const int N = 5e6+5;
double h[505][10005],res[505][10005];
bool vis[505][10005];
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
int n,m;
struct node
{
double h;
int x,y;
};
vector<node>po;
bool check(int x,int y)
{
return x>0 and x<=n and y>0 and y<=n;
}
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
cin>>h[i][j];
res[i][j] = (double)m;
po.push_back({h[i][j],i,j});
}
sort(po.begin(),po.end(),[](node a,node b){
return a.h > b.h;
});
for(int i=0;i<po.size();i++)
{
int cnt = 0;
int xx = po[i].x,yy = po[i].y;
for(int j=0;j<4;j++)
{
int nx = xx+dx[j];
int ny = yy+dy[j];
if(!check(nx,ny)) continue;
if(!vis[nx][ny] and po[i].h>h[nx][ny]) cnt++;
}
for(int j=0;j<4;j++)
{
int nx = xx+dx[j];
int ny = yy+dy[j];
if(!check(nx,ny)) continue;
if(!vis[nx][ny] and po[i].h>h[nx][ny])
res[nx][ny] += res[xx][yy]/cnt;
}
vis[xx][yy] = true;
}
for(int i=1以上是关于2021 ICPC 网络赛2的主要内容，如果未能解决你的问题，请参考以下文章
2021ICPC网络赛第二场The 2021 ICPC Asia Regionals Online Contest (II) L Euler Function
ICPC 2021网络赛2The 2021 ICPC Asia Regionals Online Contest (II)签到题5题