2014 北京邀请赛ABDHJ题解
A. A Matrix
点击打开链接构造，结论是从第一行開始往下产生一条曲线，使得这条区间最长且从上到下递减，
#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:点击打开链接
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：点击打开链接
先得到一个dp方程
dp[i][j]表示字符串长度为i，结尾有j个数互不同样的方法数
然后得到转移方程再用矩阵高速幂加速
#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：点击打开链接
把全部数字取法后都一起排序，最大的减最小的就可以
可能最大最小都来源一个数，则答案就是 第二大减最小 或者最大减第二小
#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：点击打开链接
hash二分第一个不同的字符距离0的位置，再二分第二个不同的字符距离第一个字符的位置
#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; }
