| A - Pens and Pencils |
emm題意忘了
| B - Rooms and Staircases |
有兩層房子，每層有k間 每間要麼是0要麼是1 0只能左右， 1可以上下 問最多能走多少間。
不用考慮上下左右來回走，枚舉每個轉折上面走最大下面走最大更新極值即可，來回走必然不如這樣優
/*
Zeolim - An AC a day keeps the bug away
*/
//#pragma GCC optimize(2)
//#pragma GCC ("-W1,--stack=128000000")
#include <bits/stdc++.h>
using namespace std;
#define mp(x, y) make_pair(x, y)
#define fr(x, y, z) for(int x = y; x < z; ++x)
#define pb(x) push_back(x)
#define mem(x, y) memset(x, y, sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <int, int> pii;
typedef std::vector <int> vi;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
const ull P = 13331;
const int MAXN = 2e6 + 10;
int dp[MAXN][2] = {0};
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
//freopen("1.txt", "r", stdin);
int T;
cin >> T;
while(T--)
{
int n;
cin >> n;
int ans = 0;
string s;
cin >> s;
if(s.find('1') == string::npos)
cout << n << '\n';
else
{
int cnt = 0;
int fst = INF;
int lst = -INF;
for(int i = 0; i < n; ++i)
{
if(s[i] == '1')
{
++cnt;
fst = min(fst, i);
lst = max(lst, i);
}
}
++fst, ++lst;
int ans = 0;
ans = max(ans, n - fst + 1);
ans = max(ans, fst);
ans = max(ans, n - lst + 1);
ans = max(ans, lst);
cout << ans * 2 << '\n';
}
}
return 0;
}
| D - Paint the Tree |
給定一顆樹，相鄰三個節點顏色不能塗一樣，每個節點可以塗三個顏色，不同顏色的權值給定
首先發現，樹深度大於2必然有矛盾，所以這個樹就是鏈，然後在鏈上dp就行
其實不用dp 直接枚舉開始兩個節點的情況暴力跑6遍就行 因爲限定前兩個以後後面情況是唯一的
/*
Zeolim - An AC a day keeps the bug away
*/
//#pragma GCC optimize(2)
//#pragma GCC ("-W1,--stack=128000000")
#include <bits/stdc++.h>
using namespace std;
#define mp(x, y) make_pair(x, y)
#define fr(x, y, z) for(int x = y; x < z; ++x)
#define pb(x) push_back(x)
#define mem(x, y) memset(x, y, sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <int, int> pii;
typedef std::vector <int> vi;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f3f3f3f;
const ll MOD = 1e9 + 7;
const ull P = 13331;
const int MAXN = 1e6 + 10;
vector <int> edge[MAXN];
ll v[MAXN][3];
vector <int> ans, t;
ll rs = 0, rans = INF;
bitset <MAXN> used;
void dfs(int now, int fa, int lfa)
{
used[now] = 1;
for(int i = 0; i < 3; ++i)
{
if(i != t[fa] && i != t[lfa])
{
t[now] = i;
rs += v[now][i];
}
}
for(auto to : edge[now])
{
if(!used[to])
dfs(to, now, fa);
}
}
void gao(int beg, int x, int y)
{
used.reset();
rs = 0;
t[beg] = x;
int bto = edge[beg][0];
t[bto] = y;
rs = v[beg][x] + v[bto][y];
used[beg] = 1;
used[bto] = 1;
for(auto to : edge[bto])
{
if(!used[to])
{
dfs(to, bto, beg);
}
}
if(rs < rans)
{
rans = rs;
ans = t;
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
//freopen("1.txt", "r", stdin);
int n;
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> v[i][0];
for(int i = 1; i <= n; ++i)
cin >> v[i][1];
for(int i = 1; i <= n; ++i)
cin >> v[i][2];
for(int i = 1; i < n; ++i)
{
int x, y;
cin >> x >> y;
edge[x].pb(y);
edge[y].pb(x);
}
int beg = 0;
for(int i = 1; i <= n; ++i)
{
if(edge[i].size() > 2)
{
cout << "-1\n";
return 0;
}
if(edge[i].size() == 1)
beg = i;
}
ans.resize(n + 1);
t.resize(n + 1);
gao(beg, 0, 1);
gao(beg, 1, 0);
gao(beg, 1, 2);
gao(beg, 2, 1);
gao(beg, 0, 2);
gao(beg, 2, 0);
cout << rans << '\n';
for(int i = 1; i <= n; ++i)
cout << ans[i] + 1 << ' ';
return 0;
}
| E - Minimizing Difference |
題意：給定一個序列和一個k，你可以操作至多k次，每次操作可以將任意一個數字加1也可以將任意一個數字減1
求操作後序列最大值減最小值差值最小；
思路：首先考慮將數組中個某個值當作上界或下界，再分別二分查找以當前值爲上界/下界所能達到的最優值，更新答案即可
對於每次check，小於左邊界的所有值必須要加左邊界，大於二分右邊界的都要減到右邊界，可以在序列中二分位置+前綴和快速計算
複雜度O(nloglogn)
/*
Zeolim - An AC a day keeps the bug away
*/
//#pragma GCC optimize(2)
//#pragma GCC ("-W1,--stack=128000000")
#include <bits/stdc++.h>
using namespace std;
#define mp(x, y) make_pair(x, y)
#define fr(x, y, z) for(int x = y; x < z; ++x)
#define pb(x) push_back(x)
#define mem(x, y) memset(x, y, sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <int, int> pii;
typedef std::vector <int> vi;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 386910137;
const ull P = 13331;
const int MAXN = 1e6 + 10;
ll n, k;
ll arr[MAXN] = {0}, pre[MAXN] = {0}, rpre[MAXN] = {0};
ll rk;
bool check(ll val)
{
int pos = lower_bound(arr, arr + n, val) - arr;
ll rx = rpre[pos] - val * (n - pos);
return rx <= rk;
}
bool check1(ll val)
{
if(arr[0] >= val)
return 1;
int pos = lower_bound(arr, arr + n, val) - arr;
while(arr[pos] >= val) --pos;
ll rx = val * (pos + 1) - pre[pos];
return rx <= rk;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
//freopen("d:\out.txt","w",stdout);
//freopen("d:\in.txt","r",stdin);
cin >> n >> k;
for(ll i = 0; i < n; ++i)
cin >> arr[i];
sort(arr, arr + n);
pre[0] = arr[0];
for(ll i = 1; i < n; ++i)
pre[i] = pre[i - 1] + arr[i];
rpre[n - 1] = arr[n - 1];
for(ll i = n - 2; i >= 0; --i)
rpre[i] = rpre[i + 1] + arr[i];
ll ans = INF;
for(ll i = 0; i < n; ++i)
{
ll rval = arr[i] * (i + 1);
ll rx = rval - pre[i];
if(rx > k) break;
rk = k - rx;
ll fst = arr[i], lst = arr[n - 1], mid;
while(fst <= lst)
{
mid = (fst + lst) / 2;
if(check(mid))
{
ans = min(ans, mid - arr[i]);
lst = mid - 1;
}
else
{
fst = mid + 1;
}
}
}
for(ll i = n - 1; i >= 0; --i)
{
ll rval = arr[i] * (n - i);
ll rx = rpre[i] - rval;
if(rx > k) break;
rk = k - rx;
ll fst = 0, lst = arr[i], mid;
while(fst <= lst)
{
mid = (fst + lst) / 2;
if(check1(mid))
{
ans = min(ans, arr[i] - mid);
fst = mid + 1;
}
else
{
lst = mid - 1;
}
}
}
cout << ans << '\n';
return 0;
}