标签:中间 现在 precision pen scan size add 方向 问题
Solved | A | B | C | D | E | F | G | H | I | J | K |
---|---|---|---|---|---|---|---|---|---|---|---|
8 / 11 | ? | O | - | - | ? | - | ? | O | O | ? | O |
题意:
二维平面中给定\(n\)个点\((x_i,y_i)\)。
现在要在二维平面中构造\(n\)个点,其中第\(i\)个点与\((x_i,y_i)\)相连,构造的点可以重合,贡献为边权的平方。假设构造的第\(i\)个点为\((X_i,Y_i)\),那么对于\(i<j,X_i\leq X_j,Y_i\leq Y_j\)。
问最后最小的贡献为多少。
思路:
其实就是经典的一个idea的变形,但可以把序列变为连续的若干段,这样问题就变为经典题目了。
详见代码:
/*
* Author: heyuhhh
* Created Time: 2020/6/10 14:55:43
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#include <stack>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
void run() {
int n; cin >> n;
vector <int> x(n), y(n);
for (int i = 0; i < n; i++) {
cin >> x[i] >> y[i];
}
auto gao = [&] (vector <int>& a) {
stack <pair <ll, int>> s;
for (int i = 0; i < n; i++) {
pair <ll, int> now = MP(a[i], 1);
while (!s.empty() && s.top().fi * now.se > now.fi * s.top().se) {
pair <ll, int> t = s.top(); s.pop();
now.fi += t.fi;
now.se += t.se;
}
s.push(now);
}
double res = 0.0;
int id = n - 1;
while (!s.empty()) {
pair <ll, int> t = s.top(); s.pop();
double ave = 1.0 * t.fi / t.se;
for (int i = id; i > id - t.se; i--) {
res += (a[i] - ave) * (a[i] - ave);
}
id -= t.se;
}
return res;
};
double ans = 0.0;
ans += gao(x);
ans += gao(y);
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
拓扑序+贪心。
从前往后贪心正确性得不到保障,但是从后往前即可解决这一问题。
很有意思,拓扑序有时候反着来就能解决很多问题QAQ。
代码写着较为复杂。。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
const int MAXN = 4e5 + 5, MAXM = 4e5 + 5, BOUND = 2e5, MOD = 1e9+7, INF = 0x3f3f3f3f, base = 10000;
const int inv2 = (MOD + 1) >> 1;
const ll INFL = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0), eps = 1e-9;
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc(x) ch[x][0]
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pair<int,int>>
#define rc(x) ch[x][1]
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define all(a) (a).begin(), (a).end()
#define sz(a) int(a.size())
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fi first
#define se second
#define MP std::make_pair
#define ri register int
//#define sz(a) int((a).size())
const int N = 2e5,M = (1<<20);
inline int add(int a, int b) {return a + b >= MOD ? a + b - MOD : a + b;}
inline int dec(int a, int b) {return a < b ? a - b + MOD : a - b;}
inline int mul(int a, int b) {return 1ll * a * b % MOD;}
template <typename T>
inline void cmin(T &a,T b){a = min(a,b);}
template <typename T>
inline void cmax(T &a,T b){a = max(a,b);}
ll qpow(ll a,ll b){
ll ans=1;
for(;b;b>>=1,a=a*a%MOD)if(b&1)ans=ans*a%MOD;
return ans;
}
mt19937 mrand(random_device{}());
int a[MAXN],f[MAXN],in[MAXN];
struct cmp{
bool operator ()(int u,int v){
return f[u] < f[v];
}
};
vi G[MAXN];
bool vis[MAXN];
void dfs(int u){
vis[u] = 1;
f[u] = a[u];
for(int v:G[u]){
if(!vis[v])dfs(v);
f[u] = max(f[u], f[v]+1);
}
}
void run(){
int n; cin>>n;
rep(i,1,n){
cin>>a[i];
int x; cin>>x;
rep(j,1,x){
int u; cin>>u;
G[u].push_back(i);
in[i]++;
}
}
rep(i,1,n){
if(!vis[i]){
dfs(i);
}
}
priority_queue<int,vector<int>, cmp> q;
rep(i,1,n)if(!in[i])q.push(i);
int ans=0,cnt=0;
while(!q.empty()){
int u = q.top();q.pop();
ans = max(ans, cnt + a[u]);
for(int v:G[u]){
if((--in[v])==0){
q.push(v);
}
}
cnt++;
}
cout<<ans<<‘\n‘;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
//freopen("A.in","r",stdin);
//freopen("A.out","w",stdout);
int _=1;
while(_--)run();
return 0;
}
题意:
给定两个表达式,每个表达式有"combine","sorted","shuffle","[]"这四种运算符。其中"[]"中间有一个序列,形如\(x_1,x_2,\cdots,x_i\)。
现在问这两个表达式是否等价。表达式等价的定义为出现每一种序列的概率都相同。
思路:
首先有几个观察:
根据这三点很容易看两个表达式最后是否相等了,但是题目要求出现的概率相等。
那么再加一个条件,在满足数列相等后,所有shuffle的区间都相等就行(此时概率肯定相等)。如果存在一个区间不相等,那么最后两个式子肯定不等价。
详见代码:
/*
* Author: heyuhhh
* Created Time: 2020/6/9 23:22:06
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
typedef pair <vector <int>, vector<pii>> result;
result get(string& s) {
int n = s.length();
vector <int> a, sum(n, -1);
for (int i = 0, j; i < n; i = j + 1) {
j = i;
if (s[i] >= ‘0‘ && s[i] <= ‘9‘) {
int x = 0;
while (j < n && s[j] >= ‘0‘ && s[j] <= ‘9‘) {
x = x * 10 + (s[j] - ‘0‘);
++j;
}
sum[j - 1] = sz(a);
a.push_back(x);
}
}
vector <pii> seg;
for (int i = 0, j; i < n; i = j + 1) {
j = i;
if (s[i] != ‘s‘) continue;
int cnt = 0;
while (j < n && (s[j] < ‘0‘ || s[j] > ‘9‘)) {
if (s[j] == ‘(‘) ++cnt;
++j;
}
int L = INF, R = -1;
while (j < n && cnt > 0) {
if (s[j] == ‘(‘) ++cnt;
if (s[j] == ‘)‘) --cnt;
if (sum[j] != -1) {
L = min(L, sum[j]);
R = max(R, sum[j]);
}
++j;
}
if (L > R) continue;
sort(a.begin() + L, a.begin() + R + 1);
if (s[i + 1] == ‘h‘ && a[L] != a[R]) {
seg.push_back(MP(L, R));
}
}
return MP(a, seg);
}
void run() {
string s, t; cin >> s >> t;
result A = get(s), B = get(t);
if (A == B) {
cout << "equal" << ‘\n‘;
} else {
cout << "not equal" << ‘\n‘;
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
题意:
给定一些操作符"LRUD",表示往某一个方向走直到遇到障碍。
现在给定终点\((0,0)\),要求构造一个起点以及若干障碍物,使得最后小球能够到达终点。注意必须是最后一步到达终点,中间到达终点即不合法。
大概图就是这个样子:
思路:
一般这种题都会考虑从后往前构造,但这个题中从后往前构造可能会很麻烦。
注意到我们终点的具体位置不重要,假设我们最后到达了某个点\((x,y)\),直接将坐标轴相对于\((x,y)\)平移即可。
所以问题就转化为了从\((0,0)\)出发,构造一种方案,使得存在一种合法的路径。
到了这一步就比较简单了。网格图中的移动一般是规定一个范围,每一次范围会变大或减小。因为这个题从\((0,0)\)开始走,所以我们不断将范围变大并且每次走到边界即可,只要保证小球不会在路径上碰到障碍物就行。
注意特殊情况:LRL,RLR,UDU,DUD出现在了末尾则肯定不合法,若中间为LRLRLR这种则不扩大范围。
细节见代码:
/*
* Author: heyuhhh
* Created Time: 2020/6/9 15:25:32
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 20 + 5;
int n;
string s;
int d;
map <int, map<int, int>> mp;
void sayNo() {
cout << "impossible" << ‘\n‘;
exit(0);
}
void run() {
cin >> s;
n = s.length();
if (n >= 3 && s[n - 1] == s[n - 3]) {
if (s[n - 1] == ‘L‘ && s[n - 2] == ‘R‘) sayNo();
if (s[n - 1] == ‘R‘ && s[n - 2] == ‘L‘) sayNo();
if (s[n - 1] == ‘U‘ && s[n - 2] == ‘D‘) sayNo();
if (s[n - 1] == ‘D‘ && s[n - 2] == ‘U‘) sayNo();
}
d = 1;
int x = 0, y = 0;
vector <pii> ans;
for (int i = 0; i < n; i++) {
if (s[i] == ‘R‘) {
x = d;
ans.push_back(MP(x + 1, y));
}
if (s[i] == ‘L‘) {
x = -d;
ans.push_back(MP(x - 1, y));
}
if (s[i] == ‘D‘) {
y = -d;
ans.push_back(MP(x, y - 1));
}
if (s[i] == ‘U‘) {
y = d;
ans.push_back(MP(x, y + 1));
}
if (i + 1 < n) {
if (s[i] == ‘L‘ && s[i + 1] == ‘R‘) {}
else if (s[i] == ‘R‘ && s[i + 1] == ‘L‘) {}
else if (s[i] == ‘U‘ && s[i + 1] == ‘D‘) {}
else if (s[i] == ‘D‘ && s[i + 1] == ‘U‘) {}
else d += 2;
}
}
cout << -x << ‘ ‘ << -y << ‘\n‘;
sort(all(ans));
ans.erase(unique(all(ans)), ans.end());
cout << sz(ans) << ‘\n‘;
for (auto it : ans) {
cout << it.fi - x << ‘ ‘ << it.se - y << ‘\n‘;
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
贪心一下就行。
从后往前插入\(1\),我们首先想让他贡献\(2\)次,如果是偶数就很简单,奇数会多一次我们插在\(1\)位置即可。
/*
* Author: heyuhhh
* Created Time: 2020/6/9 13:31:53
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
void run() {
int n, c, b;
cin >> n >> c >> b;
vector <int> a(n, -1);
for (int i = 0; i < b; i++) {
int x; cin >> x; --x;
a[x] = 0;
}
int t = c / 2;
for (int i = n - 2; i >= 0 && t; i--) {
if (a[i] == -1) {
a[i] = 1;
--t; --i;
}
}
if (c & 1) {
a[0] = 1;
}
for (int i = 0; i < n; i++) {
if (a[i] == -1) a[i] = 0;
cout << a[i];
}
cout << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
签到。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
const int MAXN = 2e5 + 5, MAXM = 4e5 + 5, BOUND = 2e5, MOD = 1e9+7, INF = 0x3f3f3f3f, base = 10000;
const int inv2 = (MOD + 1) >> 1;
const ll INFL = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0), eps = 1e-9;
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc(x) ch[x][0]
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pair<int,int>>
#define rc(x) ch[x][1]
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define all(a) (a).begin(), (a).end()
#define sz(a) int(a.size())
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fi first
#define se second
#define MP std::make_pair
#define ri register int
//#define sz(a) int((a).size())
const int N = 2e5,M = (1<<20);
inline int add(int a, int b) {return a + b >= MOD ? a + b - MOD : a + b;}
inline int dec(int a, int b) {return a < b ? a - b + MOD : a - b;}
inline int mul(int a, int b) {return 1ll * a * b % MOD;}
template <typename T>
inline void cmin(T &a,T b){a = min(a,b);}
template <typename T>
inline void cmax(T &a,T b){a = max(a,b);}
ll qpow(ll a,ll b){
ll ans=1;
for(;b;b>>=1,a=a*a%MOD)if(b&1)ans=ans*a%MOD;
return ans;
}
mt19937 mrand(random_device{}());
int c[MAXN];
void run(){
int n; cin>>n;
rep(i,1,n)cin>>c[i];
sort(c+1,c+1+n);
db frac = 1;
rep(i,1,n){
if(c[i]>i){
cout<<"-1\n";
return ;
}
frac = min(frac,(db)c[i]/i);
}
printf("%.6f\n",frac);
}
int main() {
//ios::sync_with_stdio(false); cin.tie(0);
int _=1;
while(_--)run();
return 0;
}
假设当前除开第一个人其余分数最大值为\(MAX\),有\(cnt\)个。并且现在最大和次大的差值为\(d\)。
显然模拟一下会发现经过\(log_2\lfloor\frac{cnt}{2}\rfloor\)次过后\(MAX\)会加一,并且再经过一次\(d\)会减少\(1\)。
也就是一个过程我们可以看作\(log_2\lfloor\frac{cnt}{2}\rfloor+1\)步,对于每一个值来说会重复\(d\)次,之后会更新\(MAX,cnt\)并且继续执行以上操作直到\(MAX\)超过\(a[0]\)。
那么手动模拟一下这个过程即可,注意一下边界的情况。
/*
* Author: heyuhhh
* Created Time: 2020/6/9 22:17:52
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
void run() {
int n; cin >> n;
vector <ll> a(n);
vector <int> lg(N);
lg[2] = 1;
for (int i = 3; i < N; i++) {
lg[i] = lg[i >> 1] + 1;
}
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(all(a));
reverse(all(a));
int t = 1;
ll ans = 0, Max = a[1];
while (1) {
while (t + 1 < n && a[t + 1] == a[t]) ++t;
if (t == n - 1) {
int k = lg[t];
ll D = a[0] - Max;
ll A = D / k;
Max += A * k;
ans += 1ll * (k + 1) * A;
ans += D % k;
break;
} else {
ll d = a[t] - a[t + 1];
int k = lg[t]; //+=k
if (1ll * k * d + Max <= a[0]) {
Max += 1ll * k * d;
ans += 1ll * (k + 1) * d;
} else {
ll D = a[0] - Max;
ll A = D / k;
Max += A * k;
ans += 1ll * (k + 1) * A;
ans += D % k;
break;
}
}
++t;
}
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
签到。
#include<cstdio>
using namespace std;
char enc[107];
char orn[107];
int main() {
int n, m; scanf("%d%d", &n, &m);
scanf("%s%s", orn+m-n, enc);
for(int i=m-n-1, j=m-1; i>=0; i--, j--) {
orn[i]=(enc[j]-orn[j]+26)%26+‘a‘;
}
printf("%s\n", orn);
}
2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)
标签:中间 现在 precision pen scan size add 方向 问题
原文地址:https://www.cnblogs.com/heyuhhh/p/13089638.html