2018 TCO Algorithm Round 1B
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随便搞
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 class LineOff { 5 public: 6 int movesToDo(string points) { 7 stack<char> st; 8 int ret = 0, l = points.length(); 9 for(int i = 0; i < l; ++i){ 10 if(!st.empty() && st.top() == points[i]) ret++, st.pop(); 11 else st.push(points[i]); 12 } 13 return ret; 14 } 15 };
随便搞
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef pair<int, int> pii; 4 int f[111][111][111]; 5 pii pre[111][111][111]; 6 7 class StablePairsDiv1 { 8 public: 9 void get(int step, int i, int j, vector<int> & ret){ 10 if(step != 1) get(step - 1, pre[step][i][j].first, pre[step][i][j].second, ret); 11 ret.push_back(i); 12 ret.push_back(j); 13 } 14 vector<int> findMaxStablePairs(int n, int c, int k) { 15 memset(f, -1, sizeof(f)); 16 for(int i = 1; i <= n; ++i) 17 for(int j = i + 1; j <= n; ++j) 18 f[1][i][j] = i + j; 19 for(int i = 1; i < k; ++i){ 20 for(int j = 1; j <= n; ++j){ 21 for(int k = j + 1; k <= n; ++k){ 22 if(f[i][j][k] == -1) continue; 23 for(int p = k + 1; p <= n; ++p){ 24 int q = j + k + c - p; 25 if(q <= p) break; 26 int t = f[i][j][k] + p + q; 27 if(t > f[i+1][p][q]) f[i+1][p][q] = t, pre[i+1][p][q] = pii(j, k); 28 } 29 } 30 } 31 } 32 int ans = -1, ii, jj; 33 for(int i = 1; i <= n; ++i){ 34 for(int j = i + 1; j <= n; ++j){ 35 if(f[k][i][j] > ans) ans = f[k][i][j], ii = i, jj = j; 36 } 37 } 38 vector<int> ret; 39 if(ans == -1) return ret; 40 get(k, ii, jj, ret); 41 return ret; 42 } 43 };
随便搞
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const LL mod = 1e9 + 7; 5 6 LL f[66][66][4][2]; 7 class ThreeSameLetters { 8 public: 9 int countStrings(int L, int S){ 10 for(int i = 1; i <= S; ++i) f[1][i][1][0] = 1; 11 for(int i = 1; i < L; ++i){ 12 for(int j = 1; j <= S; ++j){ 13 for(int k = 1; k <= 3; ++k){ 14 for(int p = 0; p <= 1; ++p){ 15 for(int q = 1; q <= S; ++q){ 16 int nk = q == j ? k + 1 : 1; 17 if(nk > 3) continue; 18 if(p && nk == 3) continue; 19 int np = p || nk == 3; 20 f[i+1][q][nk][np] = (f[i+1][q][nk][np] + f[i][j][k][p]) % mod; 21 } 22 } 23 } 24 } 25 } 26 LL ans = 0; 27 for(int i = 1; i <= S; ++i) ans = (ans + f[L][i][1][1] + f[L][i][2][1] + f[L][i][3][1]) % mod; 28 return (int) ans; 29 } 30 };
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