Codeforces Round 918 (Div. 4)（AK）

A、模拟

B、模拟

C、模拟

D、模拟

E、思维，前缀和

F、思维、逆序对

G、最短路

A - Odd One Out

题意：给定三个数字，有两个相同，输出那个不同的数字。

直接傻瓜写法

void solve() 
{int a , b , c;cin >> a >> b >> c;if(a == b){cout << c << endl;}	else if(a == c){cout << b << endl;}elsecout << a << endl;
}

B - Not Quite Latin Square

题意：给定一个3*3的矩阵，每一行每一列都有且仅有A、B、C三个字母组成。现在给出矩阵中有一个? , 求这个？代表哪个字母。

可以用二进制表示三个字母是否存在

void solve() 
{string s[3];for(int i = 0 ; i < 3; i ++)cin >> s[i];for(int i = 0 ; i < 3 ;i ++){int mask = 0;for(int j = 0 ; j < 3 ; j ++){mask += (1 << (s[i][j] - 'A')) * (s[i][j] != '?');}if(mask != 7){for(int j = 0 ; j < 3 ; j ++){if((mask >> j) & 1){continue; }else{char c = j + 'A';cout << c <<endl;}}}}
}

C - Can I Square?

题意：给定一个数组，求数组之和能否形成完全平方数。

注意：直接用sqrt会因为精度问题而出错，所以在[sqrt - 2 , sqrt + 2]之间都试一遍即可。

void solve() 
{LL sum = 0;cin >> n;for(int i = 0 ; i < n ; i ++){int x;cin >> x;sum += x;}	LL t = sqrt(sum);for(LL i = t - 1 ; i <= t + 1 ; i ++){if(i < 0)continue;if(i * i == sum){cout <<"YES\n";return;}}cout <<"NO\n";
}

D - Unnatural Language Processing

题意：

思路：将a、e看成0，b、c、d看成1。整个单词变成了一个01串，然后发现：当连续的两个1出现时，前一个1需要放到前面的音节结尾。当只有一个连续的1，那么这个1就是音节的开头。然后模拟整个过程就行。

// Problem: D. Unnatural Language Processing
// Contest: Codeforces - Codeforces Round 918 (Div. 4)
// URL: https://codeforces.com/contest/1915/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
//10 101
void solve() 
{cin >> n;string s;cin >> s;for(int i = 0 ; i < n ; i ++){if(s[i] == 'a' || s[i] == 'e'){a[i] = 0;}else{a[i] = 1;}}for(int i = 0 ; i < n ; i ++){if(a[i] == 1){cout << s[i];}else if(a[i] == 0){if(i < n - 3){if(a[i + 1] == 1 && a[i + 2] == 1){cout << s[i] << s[i + 1] <<"."; i++;}else{cout << s[i] <<".";}}else if(i == n - 3){cout << s[i] <<".";}else if(i == n - 2){cout << s[i] << s[i + 1];i++;}else{cout << s[i];}}}cout << endl;
}            
int main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

E - Romantic Glasses

题意：给定一数组，求其中是否存在某个连续子序列是否满足 $\sum _{i = l}^{r}a[i](i\ mod\ 2==0) = \sum _{i = l}^{r}a[i](i\ mod\ 2 == 1)$

思路：转移之后有公式 $\sum _{i = l}^{r}a[i](i\ mod\ 2==0) + \sum _{i = l}^{r}-a[i](i\ mod\ 2 == 1) = 0$ ，也就是对于原数组的奇数项都乘 -1 之后，求是否存在某个区间之和为0。

用前缀和 $sum$ 数组来表示区间，也就是存在 $sum(r) - sum(l) = 0$ , 也就是 $sum(r) = sum(l)$ 。因此我们可以逐步递增 $r$ ，然后看之前是否出现过 $sum(l)$ 与 $sum(r)$ 相等。可用map或者set来存之前出现过的前缀和情况。这样整个复杂度为 $O(NlogN)$

// Problem: E. Romantic Glasses
// Contest: Codeforces - Codeforces Round 918 (Div. 4)
// URL: https://codeforces.com/contest/1915/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{cin >> n;for(int i = 0 ; i < n ; i ++){cin >> a[i];if(i % 2 == 1){a[i] *= -1;}}	set<int>pre;pre.insert(0);int sum = 0;for(int i = 0 ; i < n ; i ++){sum += a[i];//cout << sum << endl;if(pre.count(sum)){cout <<"YES\n";return;}pre.insert(sum);}cout <<"NO\n";return;
}            
signed main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

F - Greetings

题意：

思路：将所有人的起点按照从小到大进行排序，这样就满足了后面的人不会撞到前面的人（只存在后面的人已经到达终点了，然后被前面的人撞）。然后再考虑能够撞多少个人：对于排完序以后的第 $i$ 个人而言，他能撞到的人是 $i$ 以后的，终点小于等于 $b_{i}$ 的人。因此也就是 $(b_{i} \geq b_{j})(i < j)$ 的个数，也就是按照起点排完序之后的 $b$ 数组的逆序对数量，然后套一遍逆序对的板子即可。

// Problem: F. Greetings
// Contest: Codeforces - Codeforces Round 918 (Div. 4)
// URL: https://codeforces.com/contest/1915/problem/F
// Memory Limit: 256 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
int tmp[N];
LL merge_sort(int q[], int l, int r)
{if (l >= r) return 0;int mid = (l + r) >> 1; // 二分区间LL res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);//归并int i = l, j = mid + 1, k = 0;while (i <= mid && j <= r){if (q[i] <= q[j]) tmp[k ++] = q[i ++]; // 前面的排序正常，注意`=` 说明不是逆序对else{res += mid - i + 1;tmp[k ++] = q[j ++];}}// 扫尾工作while (i <= mid) tmp[k ++] = q[i ++];while (j <= r) tmp[k ++] = q[j ++];for (int i = l, j = 0; i <= r; i ++ , j ++) q[i] = tmp[j];return res;
}
void solve() 
{cin >> n;pair<int,int>po[n];for(int i = 0 ; i < n ; i++){cin >> po[i].x >> po[i].y;}	sort(po , po + n);int a[n];for(int i = 0 ; i < n ; i ++){a[i] = po[i].y;}LL ans = merge_sort(a , 0 , n  - 1);cout << ans <<endl;
}            
int main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

G - Bicycles

题意：

思路：观察到数据不大。因此直接考虑最短路算法。需要注意的是，整个过程不仅仅有点这一个条件，还有自行车速度系数这个限制。因此需要将这两个限制都表示出来，具体看代码注释

// Problem: G. Bicycles
// Contest: Codeforces - Codeforces Round 918 (Div. 4)
// URL: https://codeforces.com/contest/1915/problem/G
// Memory Limit: 256 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 1010;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
struct node{int num;int dis;bool operator > (const node &t) const{return dis > t.dis;}int own;
}tmp;
int dis[N][N];//到达i点，且拥有自行车系数j的最短距离
int vis[N][N];//到达i点，且拥有自行车系数j的可能性
vector<node>tr[N];
int cost[N];
void dij(int s)
{for(int i = 1 ; i <= n ; i ++){for(int j = 0 ; j <= 1000 ; j ++){dis[i][j] = llinf;}}priority_queue<node,vector<node> , greater<node> > q;q.push({s , 0 , cost[1]});dis[s][cost[1]] = 0;while(!q.empty()){tmp = q.top();int x = tmp.num;//所在地int y = tmp.own;//拥有的自行车系数q.pop();if(vis[x][y] == 1)continue;vis[x][y] = 1;for(int i = 0 ; i < (int)tr[x].size() ; i ++ ){node now = tr[x][i];int len = tr[x][i].dis;//距离int e = tr[x][i].num;//目标地int pp = min(y , cost[e]);//到达目的地之后所拥有的自行车系数if(dis[e][pp] > dis[x][y] + len * y){dis[e][pp] = dis[x][y] + len * y;q.push({e , dis[e][pp] , pp});}}}
}
void solve() 
{cin >> n >> m;for(int i = 1 ; i <= n ; i ++){tr[i].clear();for(int j = 0 ; j <= 1000 ; j ++){vis[i][j] = 0;}}for(int i = 0 ; i < m ; i ++){int u , v , dis;cin >> u >> v >> dis;tr[u].pb({v , dis , 0});tr[v].pb({u , dis , 0});}for(int i = 1 ; i <= n ; i ++){cin >> cost[i];}dij(1);int ans = llinf;for(int i = 0 ;i <= 1000 ;  i++){ans = min(ans , dis[n][i]);}cout << ans << endl;
}            
signed main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}