2014 北京邀请赛ABDHJ题解
A. A Matrix
點擊打開鏈接構造，結論是從第一行開始往下產生一條曲線，使得這條區間最長且從上到下遞減，
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <stdio.h> #include <vector> #include <set> using namespace std; #define N 100005 vector<int>G[N], P[N], tmp; set<int>s[N]; set<int>::iterator it; int n,m; bool work(){for(int i = m; i>1; i--) { for(int j = G[i].size()-1; j>=0; j--) {it = s[i-1].lower_bound(-G[i][j]);if(it==s[i-1].end())return false;int pos = lower_bound(G[i-1].begin(), G[i-1].end(), -(*it))-G[i-1].begin();P[i-1][pos] = j; s[i-1].erase(it);}}return true; } void init(){for(int i = 0; i <= n; i++)G[i].clear(), s[i].clear(), P[i].clear(); } int Stack[N]; int main(){int T, Cas = 1,i,j,k;cin>>T;while(T--) {init();scanf("%d %d",&n,&m);bool success = true;for(i=1;i<=m;i++){scanf("%d",&k);for(int z = 0; z < k; z++){scanf("%d",&j);G[i].push_back(j);s[i].insert(-j);if(z && G[i][z-1]>G[i][z]) success=false;P[i].push_back(-1);}}printf("Case #%d: ",Cas++);if(success && work()) {int top = 0;for(i = 0; i < P[1].size(); i++){ int h = 1, l = i;tmp.clear();while(1){tmp.push_back(G[h][l]);if(P[h][l]==-1)break;l = P[h][l];h++; } for(j=tmp.size()-1;j>=0;j--)Stack[top++] = tmp[j]; }for(i=0;i<top;i++)printf("%d%c",Stack[i],i==top-1?'\n':' ');}else puts("No solution");}return 0; }B. Beautiful Garden
B:點擊打開鏈接
暴力
#include <cstdio> #include <cstring> #include <algorithm> typedef long long ll; const int N = 42; int n, a[N]; int dis[N * N], dep; bool vis[N][N]; int main() {int T;scanf("%d", &T);for (int cas = 1; cas <= T; ++cas) {scanf("%d", &n);for (int i = 0; i < n; ++i)scanf("%d", &a[i]);std::sort(a, a + n);dep = 0;for (int i = 0; i < n; ++i)for (int j = i + 1; j < n; ++j)dis[dep++] = a[j] - a[i];std::sort(dis, dis + dep);dep = std::unique(dis, dis + dep) - dis;int ans = n - 1, cnt, idx, f;ll st;for (int i = 0; i < dep; ++i) {memset(vis, 0, sizeof vis);for (int k = 0; k < n; ++k)for (int z = 0; z < n; ++z)if(!vis[k][z]) {st = a[k] - (ll)z * dis[i];cnt = idx = 0;f = 1;for (int l = 0; l < n; ++l) {while (idx < n && a[idx] < st)++idx;if (idx < n && st == a[idx]) {if (vis[idx][l]) {f = 0;break;}vis[idx][l] = true;++cnt, ++idx;}st += dis[i];}if (f && n - cnt < ans)ans = n - cnt;}}printf("Case #%d: %d\n", cas, ans);}return 0; }E. Elegant String
E：點擊打開鏈接
先得到一個dp方程
dp[i][j]表示字符串長度為i，結尾有j個數互不同樣的方法數
然后得到轉移方程再用矩陣高速冪加速
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <stdio.h> #include <vector> #include <set> using namespace std; #define ll long long #define Matr 11 #define mod 20140518 struct mat{ll a[Matr][Matr],size;mat(){size = 0;memset(a, 0, sizeof a);} }; mat multi(mat m1,mat m2){mat ans = mat(); ans.size = m1.size;for(int i = 1; i <= m1.size; i++)for(int j = 1; j<= m2.size; j++)if(m1.a[i][j])for(int k = 1; k <= m1.size; k++)ans.a[i][k] = (ans.a[i][k]+m1.a[i][j]*m2.a[j][k])%mod;return ans; } mat quickmulti(mat m,ll n){mat ans = mat();ll i;for(i=1;i<=m.size;i++)ans.a[i][i] = 1;ans.size = m.size;while(n){if(n&1)ans = multi(m,ans);m=multi(m,m);n /= 2;}return ans; } ll n,k; int main(){ll T, Cas = 1;cin>>T;while(T--){cin>>n>>k;mat z = mat();for(ll i = 1; i <= k; i++)for(ll j = i; j <= k; j++)z.a[i][j] = 1;ll now = k;for(ll i = 2; i <= k; i++, now--)z.a[i][i-1]+=now;z.size = k;z = quickmulti(z, n-1);ll tmp = 0;for(ll i = 1; i <= k; i++)tmp += z.a[i][1];tmp *= (k+1);tmp %= mod;printf("Case #%lld: %lld\n",Cas++,tmp);}return 0; }H. Happy Reversal
H：點擊打開鏈接
把全部數字取法后都一起排序，最大的減最小的就可以
可能最大最小都來源一個數，則答案就是 第二大減最小 或者最大減第二小
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int N = 20005; ll a[N], b[N]; ll er[N]; int main() {int T;char s[66];er[0] = 1;for(int i = 1;i<63;i++)er[i] = er[i-1]*2;while(~scanf("%d", &T)) {for(int cas = 1; cas <= T; cas ++) {int n, m;scanf("%d%d", &n, &m);for(int i = 0; i < n; i ++) {scanf("%s", s);ll suma = 0;ll sumb = 0;for(int j = 0; j < m; j ++) {suma = suma * 2 + (s[j] == '1');sumb = sumb * 2 + (s[j] == '0');}a[i*2] = max(suma, sumb);a[i*2+1] = min(suma, sumb);}sort(a, a + n + n);ll ans;if( a[2*n-1] + a[0] + 1 == er[m] ) ans = max(a[2*n-2] - a[0], a[2*n-1] - a[1]);else ans = a[2*n-1] - a[0];printf("Case #%d: %lld\n", cas, ans);}}return 0; }J. Justice String
J：點擊打開鏈接
hash二分第一個不同的字符距離0的位置，再二分第二個不同的字符距離第一個字符的位置
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int MAX_N = 3000007; int sa[MAX_N], ws[MAX_N], rank[MAX_N], height[MAX_N]; #define F(x) ((x)/3+((x)%3==1?0:tb)) #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) int wa[MAX_N],wb[MAX_N],wv[MAX_N],wss[MAX_N]; struct suffix { int c0(int *r, int a, int b) { return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2]; } int c12(int k,int *r,int a,int b) { if(k == 2) return r[a] < r[b] || ( r[a] == r[b] && c12(1,r,a+1,b+1) ); else return r[a] < r[b] || ( r[a] == r[b] && wv[a+1] < wv[b+1] ); } void sort(int *r,int *a,int *b,int n,int m) { int i; for(i = 0;i < n;i++)wv[i] = r[a[i]]; for(i = 0;i < m;i++)wss[i] = 0; for(i = 0;i < n;i++)wss[wv[i]]++; for(i = 1;i < m;i++)wss[i] += wss[i-1]; for(i = n-1;i >= 0;i--) b[--wss[wv[i]]] = a[i]; } void dc3(int *r,int *sa,int n,int m) { int i, j, *rn = r + n; int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p; r[n] = r[n+1] = 0; for(i = 0;i < n;i++)if(i %3 != 0)wa[tbc++] = i; sort(r + 2, wa, wb, tbc, m); sort(r + 1, wb, wa, tbc, m); sort(r, wa, wb, tbc, m); for(p = 1, rn[F(wb[0])] = 0, i = 1;i < tbc;i++) rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++; if(p < tbc)dc3(rn,san,tbc,p); else for(i = 0;i < tbc;i++)san[rn[i]] = i; for(i = 0;i < tbc;i++) if(san[i] < tb)wb[ta++] = san[i] * 3; if(n % 3 == 1)wb[ta++] = n - 1; sort(r, wb, wa, ta, m); for(i = 0;i < tbc;i++)wv[wb[i] = G(san[i])] = i; for(i = 0, j = 0, p = 0;i < ta && j < tbc;p++) sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++]; for(;i < ta;p++)sa[p] = wa[i++]; for(;j < tbc;p++)sa[p] = wb[j++]; } void DA(int str[],int sa[],int n,int m) { for(int i = n;i < n*3;i++) str[i] = 0; dc3(str, sa, n+1, m); int i,j,k = 0; for(i = 0;i <= n;i++)rank[sa[i]] = i; for(i = 0;i < n; i++) { if(k) k--; j = sa[rank[i]-1]; while(str[i+k] == str[j+k]) k++; height[rank[i]] = k; } } }; int RMQ[MAX_N], mm[MAX_N], best[20][MAX_N]; void initRMQ(int n) { mm[0] = -1; for(int i = 1; i <= n; ++i) { mm[i]=((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1]; } for (int i = 1; i <= n; ++i) { best[0][i] = i; } for (int i = 1; i <= mm[n]; ++i) { for (int j = 1; j + (1 << i) - 1 <= n; ++j) { int a = best[i - 1][j]; int b = best[i - 1][j + (1 << (i - 1))]; if (RMQ[a] < RMQ[b]) best[i][j] = a; else best[i][j] = b; } } } int query(int a, int b) { int t = mm[b - a + 1]; b -= (1 << t) - 1; a = best[t][a], b = best[t][b]; return RMQ[a] < RMQ[b] ?
a : b; } int lcp(int a, int b) { a = rank[a], b = rank[b]; if (a > b) swap(a, b); return height[query(a + 1, b)]; } char A[MAX_N],B[MAX_N]; int r[MAX_N], da[MAX_N]; int main() { int T; int cas = 0; scanf("%d",&T); while(T-- > 0) { suffix ar; scanf("%s%s",A, B); int len1 = strlen(A); int len2 = strlen(B); if(len1 < len2) { printf("Case #%d: -1\n", ++cas); continue; } int n = len1 + len2 + 1; for(int i = 0; i < len1; ++i) r[i] = A[i]; for(int i = 0; i < len2; ++i) r[len1 + 1 + i] = B[i]; r[len1] = 1; r[n] = 0; ar.DA(r, sa, n, 128); for(int i = 1; i <= n; ++i) RMQ[i] = height[i]; initRMQ(n); int ans = -1; for(int i = 0; i <= len1 - len2; ++i) { int st = i; int ed = len1 + 1; int tmp = lcp(st, ed); // printf("here1 %d %d %d\n", st, ed, tmp); st += tmp; ed += tmp; if(ed >= n) { ans = i; break; } st++; ed++; if(ed >= n) { ans = i; break; } tmp = lcp(st, ed); // printf("here2 %d %d %d\n", st, ed, tmp); st += tmp; ed += tmp; if(ed >= n) { ans = i; break; } st++; ed++; if(ed >= n) { ans = i; break; } tmp = lcp(st, ed); // printf("here3 %d %d %d\n", st, ed, tmp); st += tmp; ed += tmp; if(ed >= n) { ans = i; break; } } printf("Case #%d: %d\n", ++cas, ans); } return 0; }
轉載于:https://www.cnblogs.com/bhlsheji/p/5228348.html
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