Codeforces Round #738 (Div. 2)

文章目錄

• A
• • 題解：
  • 代碼：
• B
• • 題意：
  • 題解：
  • 代碼：
• C
• • 題意：
  • 題解：
  • 代碼：
• D1
• • 題意：
  • 題解：
  • 代碼:

題號題目知識點

A	Mocha and Math
B	Mocha and Red and Blue
C	Mocha and Hiking
D1	Mocha and Diana (Easy Version)
D2	Mocha and Diana (Hard Version)
E	Mocha and Stars

A

題解：

可以任意選區間，可以操作多次，也就是任何數都可以進行&操作，所以答案就是所有&的結果

代碼：

// Problem: A. Mocha and Math // Contest: Codeforces - Codeforces Round #738 (Div. 2) // URL: https://codeforces.com/contest/1559/problem/A // Memory Limit: 256 MB // Time Limit: 1000 ms // Data:2021-08-16 19:15:42 // By Jozky#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; template <typename T> inline void read(T& x) {T f= 1;x= 0;char ch= getchar();while (0 == isdigit(ch)) {if (ch == '-')f= -1;ch= getchar();}while (0 != isdigit(ch))x= (x << 1) + (x << 3) + ch - '0', ch= getchar();x*= f; } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin); #endif } void Time_test() { #ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } int main() {//rd_test();int t;read(t);while (t--) {int n;read(n);ll ans= 0;for (int i= 1; i <= n; i++) {int x;read(x);if (i == 1)ans= x;elseans= ans & x;}printf("%d\n", ans);}return 0;//Time_test(); }

B

題意：

長度為n的字符串，由BR？三種符號組成，現在要求你將？填上B/R，使得相同符號相鄰的情況最少

題解：

找到第一個非？的字符，然后從該點開始向前向后一次填充，每次填充？時填與其相鄰的相反元素
這樣保證最優

代碼：

// Problem: B. Mocha and Red and Blue // Contest: Codeforces - Codeforces Round #738 (Div. 2) // URL: https://codeforces.com/contest/1559/problem/B // Memory Limit: 256 MB // Time Limit: 1000 ms // Data:2021-08-16 19:24:33 // By Jozky#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; template <typename T> inline void read(T& x) {T f= 1;x= 0;char ch= getchar();while (0 == isdigit(ch)) {if (ch == '-')f= -1;ch= getchar();}while (0 != isdigit(ch))x= (x << 1) + (x << 3) + ch - '0', ch= getchar();x*= f; } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin); #endif } void Time_test() { #ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } int main() {//rd_test();int t;read(t);while (t--) {int n;read(n);string s;cin >> s;int pos= n + 1;for (int i= 0; i < s.length(); i++) {if (s[i] != '?') {pos= i;break;}}for (int i= pos - 1; i >= 0; i--) {if (s[i] == '?') {if (s[i + 1] == 'R')s[i]= 'B';elses[i]= 'R';}}for (int i= pos + 1; i < s.length(); i++) {if (s[i] == '?') {if (s[i - 1] == 'R')s[i]= 'B';elses[i]= 'R';}}cout << s << endl;}//Time_test(); }

C

題意：

有n+1個點，其中1~n個點，1有條到2的邊，2有條到3的邊…n-1有條到n的邊
1到n這些點與n的連邊關系題目給出，問能否將所有點全走一遍(每個點最多只能走一次)

題解：

不難發現，從1出發，是可以到n的(經過了1到n所有點)，現在還剩n+1
如果點n+1能到1或者n能到n+1,那就可以順利到點n+1
或者從$a_i$可以到n+1，從n+1也可以到$a_{i+1}$，這樣也行

代碼：

// Problem: C. Mocha and Hiking // Contest: Codeforces - Codeforces Round #738 (Div. 2) // URL: https://codeforces.com/contest/1559/problem/C // Memory Limit: 256 MB // Time Limit: 1000 ms // Data:2021-08-16 19:47:58 // By Jozky#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; template <typename T> inline void read(T& x) {T f= 1;x= 0;char ch= getchar();while (0 == isdigit(ch)) {if (ch == '-')f= -1;ch= getchar();}while (0 != isdigit(ch))x= (x << 1) + (x << 3) + ch - '0', ch= getchar();x*= f; } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin); #endif } void Time_test() { #ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn= 2e4 + 9; vector<int> vec[maxn]; int a[maxn]; int main() {//rd_test();int t;read(t);while (t--) {int n;read(n);for (int i= 1; i < n; i++)vec[i].push_back(i + 1);for (int i= 1; i <= n; i++) {cin >> a[i];}if (a[1] == 1) {printf("%d", n + 1);for (int i= 1; i <= n; i++) {printf(" %d", i);}printf("\n");continue;}else if (a[n] == 0) {printf("%d", 1);for (int i= 2; i <= n + 1; i++) {printf(" %d", i);}printf("\n");continue;}bool f= 0;for (int i= 1; i < n; i++) {if (a[i] == 0 && a[i + 1] == 1) {for (int j= 1; j <= n; j++) {printf("%d ", j);if (j == i)printf("%d ", n + 1);}printf("\n");break;}}if (f == 0)continue;elseprintf("-1\n");}//Time_test(); }

D1

題意：

有兩個森林，現在要求你在第一個森林加邊，第二個森林會自動進行一樣的操作，兩個森林都不允許出現環，問最多能加多少邊？

題解：

利用并查集分開維護兩個森林，對于邊u和v，如果uv在第一個森林不在一個集合內，在第二個森林也不在一個集合內，才可以加入
O($n^2$)

代碼:

// Problem: D1. Mocha and Diana (Easy Version) // Contest: Codeforces - Codeforces Round #738 (Div. 2) // URL: https://codeforces.com/contest/1559/problem/D1 // Memory Limit: 256 MB // Time Limit: 1000 ms // Data:2021-08-16 20:01:37 // By Jozky#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; template <typename T> inline void read(T& x) {T f= 1;x= 0;char ch= getchar();while (0 == isdigit(ch)) {if (ch == '-')f= -1;ch= getchar();}while (0 != isdigit(ch))x= (x << 1) + (x << 3) + ch - '0', ch= getchar();x*= f; } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin); #endif } void Time_test() { #ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn= 3000; int vis[maxn]; int fa[maxn]; int find(int x) {if (fa[x] != x)return fa[x]= find(fa[x]);return x; } void join(int u, int v) {u= find(u);v= find(v);if (u != v)fa[u]= v; } int main() {//rd_test();int n, m1, m2;cin >> n >> m1 >> m2;if (m1 == n - 1 && m2 == n - 1) {printf("0");return 0;}int tot= min(n - m1 - 1, n - m2 - 1);printf("%d\n", tot);for (int i= 1; i <= 2 * n; i++)fa[i]= i;for (int i= 1; i <= m1; i++) {int u, v;cin >> u >> v;join(u, v);}for (int i= 1; i <= m2; i++) {int u, v;cin >> u >> v;join(u + n, v + n);}// for (int i= 1; i <= n; i++) {// printf("%d ", fa[i + n]);// }// printf("\n");for (int i= 1; i <= n; i++) {for (int j= 1; j <= n; j++) {if (i == j)continue;if (find(i) != find(j)) {if (find(i + n) != find(j + n)) {printf("%d %d\n", i, j);join(i, j);join(i + n, j + n);tot--;if (tot == 0)return 0;}}}}return 0;//Time_test(); }

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