标签：ORC 贪心 version 题目 review define max style with
C. Yet Another Walking Robot
每次把走到的点记录下来，如果后续出现了一个点在之前已经出现过了，那么肯定出现了重复走。另外维护下 r-l 要尽可能小
#include <iostream> #include <algorithm> #include <string> #include <string.h> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <math.h> #include <cstdio> #include <iomanip> #include <time.h> #include <bitset> #define LL long long #define INF 0x3f3f3f3f #define ls nod<<1 #define rs (nod<<1)+1 using namespace std; const int maxn = 2e5 + 10; const LL mod = 1e9 + 7; char s[maxn]; int main() { int T; cin >> T; while (T--) { int n; cin >> n; cin >> s+1; map<pair<int,int>,int > mp; mp[make_pair(0,0)] = 1; int x = 0,y = 0; int r = 1e9,l = 0; for (int i = 1;i <= n;i++) { if(s[i]==‘L‘){ x--; }else if(s[i]==‘R‘){ x++; }else if(s[i]==‘U‘){ y++; }else { y--; } if (mp.count(make_pair(x,y))) { int l1 = mp[make_pair(x,y)]; int r1 = i; if (r1-l1 < r-l) { r = r1; l = l1; } } mp[make_pair(x,y)] = i+1; } if (l == 0) cout << -1 << endl; else cout << l << " " << r << endl; } return 0; }
D. Fight with Monsters
贪心 + 简单数学
#include <iostream> #include <algorithm> #include <string> #include <string.h> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <math.h> #include <cstdio> #include <iomanip> #include <time.h> #include <bitset> #define LL long long #define INF 0x3f3f3f3f #define ls nod<<1 #define rs (nod<<1)+1 using namespace std; const int maxn = 2e5 + 10; int v[maxn]; int main() { int n,a,b,k; int cnt = 0; cin >> n >> a >> b >> k; for (int i = 1;i <= n;i++) { cin >> v[i]; v[i] = v[i] % (a + b); if (v[i] == 0) v[i] = b; else v[i] -= a; } sort(v+1,v+1+n); for (int i = 1;i <= n;i++) { if (v[i] <= 0) { cnt++; } else { if (v[i] % a == 0) { if (k >= v[i] / a) { k -= v[i] / a; cnt++; } } else { if (k >= (v[i] / a) + 1) { k -= (v[i] / a) + 1; cnt++; } } } } cout << cnt << endl; return 0; }
E1. String Coloring (easy version)
暴力找上升的序列，然后跑两次就可以了
#include <iostream> #include <algorithm> #include <string> #include <string.h> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <math.h> #include <cstdio> #include <iomanip> #include <time.h> #include <bitset> #define LL long long #define INF 0x3f3f3f3f #define ls nod<<1 #define rs (nod<<1)+1 using namespace std; const int maxn = 2e5 + 10; const LL mod = 1e9 + 7; char s[maxn]; int vis[maxn]; int main(){ memset(vis,-1,sizeof vis); int n; cin >> n; cin >> (s + 1); char c = ‘a‘; for(int i = 1;i <= n;++i){ if(s[i] >= c ) { vis[i] = 0; c = s[i]; } } c = ‘a‘; for(int i = 1;i <= n;++i){ if(vis[i] == -1){ if(s[i] >= c) { vis[i] = 1; c = s[i]; } } } for(int i = 1;i <= n;++i) if(vis[i] == -1) { cout << "NO"; return 0; } cout << "YES" << endl; for(int i = 1;i <= n;++i){ cout << vis[i]; } }
E2. String Coloring (hard version)
题目大意：给你一串长度为n的字符串,你可以给每个位置上染上一种不大于n的颜色,对于相邻的两个位置，如果他们的颜色不同则可以交换他们的位置,现在需要交换若干次后按照字典序排序,你需要找到最少满足条件的颜色数并输出方案
可以想到，只有一个字符要与另一个字符交换，才要染成不同颜色，从前往后来考虑，我们只用考虑一个字符与在它前面的字符交换就行了，因为枚举到后面需要与其交换的字符，自然会考虑到它
只有比它大的字符才会跟它交换，所以我们只需要让它与比它大的字符不同颜色即可
然后要想到这样的一个结论，这一些字符的颜色必然在[1,??]这样的一个区间里，而且包含着[1,??]中的每一种颜色，所以我们对于枚举到的字符，找到对应的??,然后把枚举到的字符的颜色设为??+1即可
只有26个字符，暴力就完事了，复杂度??(26∗??)
#include <iostream> #include <algorithm> #include <string> #include <string.h> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <math.h> #include <cstdio> #include <iomanip> #include <time.h> #include <bitset> #define LL long long #define INF 0x3f3f3f3f #define ls nod<<1 #define rs (nod<<1)+1 using namespace std; const int maxn = 2e5 + 10; const LL mod = 1e9 + 7; int ans[maxn],maxx[maxn]; int main() { int n; cin >> n; int col = 0; for (int i = 1;i <= n;i++) { char c; cin >> c; int k = c - ‘a‘ + 1; int mx = 0; for (int j = k+1;j <= 26;j++) { mx = max(mx,maxx[j]); } ans[i] = mx + 1; col = max(col,ans[i]); maxx[k] = mx + 1; } cout << col << endl; for (int i = 1;i <= n;i++) cout << ans[i] << " "; return 0; }
F. Berland Beauty
给一棵树，边权未知，现在给m组约束，每组约束给出从u到v路径中的最小值，现在让你给出一组边权，使得符合之前的约束，不能给出输出-1
思路：
因为n较小，对于每组约束我们可以直接暴力修改路径上的权值，如果边的权值小于当前约束的最小值，则将权值修改，最后再根据每组约束暴力走一遍路径看路径是否满足要求，如果不满足则输出-1，最后还得对那些没有修改过的边随意赋值
#include <iostream> #include <algorithm> #include <string> #include <string.h> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <math.h> #include <cstdio> #include <iomanip> #include <time.h> #include <bitset> #define LL long long #define INF 0x3f3f3f3f #define ls nod<<1 #define rs (nod<<1)+1 const int maxn = 5010; const LL mod = 1e9 + 7; std::vector<int> graph[maxn]; int id[maxn][maxn]; int fa[maxn],dep[maxn]; int from[maxn],to[maxn],w[maxn]; int val[maxn][maxn]; int cst[maxn]; void dfs(int u,int f) { fa[u] = f; dep[u] = dep[f] + 1; for (int i = 0;i < graph[u].size();i++) { int v = graph[u][i]; if (v == f) continue; dfs(v,u); } } void dfs2(int u,int f) { for (int i = 0;i < graph[u].size();i++) { int v = graph[u][i]; if (v == f) continue; cst[id[u][v]] = val[u][v]; dfs2(v,u); } } int main() { int n; std::cin >> n; for (int i = 1;i <= n-1;i++) { int u,v; std::cin >> u >> v; graph[u].push_back(v); graph[v].push_back(u); id[u][v] = id[v][u] = i; } dfs(1,0); int m; std::cin >> m; for (int i = 1;i <= m;i++) { std::cin >> from[i] >> to[i] >> w[i]; if (dep[from[i]] < dep[to[i]]) std::swap(from[i],to[i]); int u = from[i],v = to[i]; while (dep[u] != dep[v]) { int f = fa[u]; if (val[u][f] <= w[i]) val[u][f] = val[f][u] = w[i]; u = f; } while (u != v) { int fu = fa[u],fv = fa[v]; if (val[u][fu] <= w[i]) val[u][fu] = val[fu][u] = w[i]; if (val[v][fv] <= w[i]) val[v][fv] = val[fv][v] = w[i]; u = fu,v = fv; } } for (int i = 1;i <= m;i++) { int x = INF; int u = from[i],v = to[i]; if (dep[u] < dep[v]) std::swap(u,v); while (dep[u] != dep[v]) { int f = fa[u]; x = std::min(x,val[u][f]); u = f; } while (u != v) { int fu = fa[u],fv = fa[v]; x = std::min(x,val[u][fu]); x = std::min(x,val[v][fv]); u = fu,v = fv; } if (x != w[i]) { std::cout << -1 << std::endl; return 0; } } dfs2(1,0); for (int i = 1;i <= n-1;i++) { if (!cst[i]) std::cout << 1000000 << " "; else std::cout << cst[i] << " "; } return 0; }
Codeforces Round #617 (Div. 3)
标签：ORC 贪心 version 题目 review define max style with
原文地址：https://www.cnblogs.com/-Ackerman/p/12317790.html
