T1 字符串构造机
考虑将一个 LCP 条件拆分成两个，一个是相等的部分，使用并查集维护，另一个是不等的部分，两个串末尾的字符一定不相等，随便那啥维护。对于非法情况就是在同一个相等联通块内有不相等的条件。然后考虑从前往后贪心即可。
#include<bits/stdc++.h>
using namespace std;
#define p pair<int, int>
constexpr int N = 1e3 + 5;
int n, m, s[N], f[N], g[N][2], tot;
inline int find(int k){ return f[k] ? f[k] = find(f[k]) : k; }
vector<int> G[N]; bitset<N> st;
int main(){
freopen("str.in", "r", stdin); freopen("str.out", "w", stdout);
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n>>m;
for(int i=1, x, y, z; i<=m; ++i){
cin>>x>>y>>z; if(x > y) swap(x, y);
int l1 = x, l2 = y, r1 = x+z-1, r2 = y+z-1;
if(x == y) continue;
if(r2+1 <= n) g[++tot][0] = r1+1, g[tot][1] = r2+1;
for(int u=l2, v=l1; u<=r2; ++u, ++v){
int fa = find(u), fb = find(v);
if(fa == fb) continue;
f[fa] = fb;
}
} memset(s, 0xff, sizeof(int) * (n+1));
for(int i=1; i<=tot; ++i){
int u = find(g[i][0]), v = find(g[i][1]);
if(u == v) return cout<<"-1", 0;
G[u].push_back(v), G[v].push_back(u);
}
for(int i=1; i<=n; ++i){
int fa = find(i);
if(s[fa] == -1){
st.reset();
for(int v : G[fa]) if(s[find(v)] != -1)
st[s[find(v)]] = 1;
int cnt = 0;
while(st[cnt]) ++cnt;
s[fa] = cnt;
} s[i] = s[fa];
}
for(int i=1; i<=n; ++i) cout<<s[i]<<' ';
return 0;
}
T2 忍者小队
最大值是简单的。找到所有 \(k\) 的倍数并取个 gcd，如果这个 gcd 不为 \(k\)，那么不存在合法的最大最小。否则最大值就为倍数的数量。
考虑最小值答案值域，一定为 \(1\sim 7\)。因为最多质数组成数最大仅为 \(2*3*5*7*11*13*17\)。那么令 \(f_{i,k}\) 表示有 \(i\) 个数 gcd 为 \(k\) 的方案数。考虑容斥，有：
\[f_{i,j}=\binom{cnt}{i}-\sum_{j=2k,j|k}^{V}f_{i,j} \]其中 \(cnt\) 为倍数数量。用方案数来 check 即可。
#include<bits/stdc++.h>
using namespace std;
constexpr int N = 3e5 + 5, M = 1e9 + 7;
int n, m, a[N], val[N], mx, f[8][N], ans[N][2], C[N][8], tot[N];
inline int add(initializer_list<int> Add){
int res = 0;
for(int v : Add) res = res + v >= M ? res + v - M : res + v;
return res;
}
int main(){
freopen("sor.in", "r", stdin); freopen("sor.out", "w", stdout);
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n>>m; for(int i=1; i<=n; ++i) cin>>a[i], ++val[a[i]], mx = max(mx, a[i]);
for(int i=0; i<=n; ++i) C[i][0] = 1;
for(int i=1; i<=n; ++i) for(int j=1; j<=7; ++j) C[i][j] = add({C[i-1][j], C[i-1][j-1]});
for(int i=1; i<=mx; ++i) ans[i][0] = 114514;
for(int k=1; k<=mx; ++k){
int tmp = 0, sum = 0;
for(int i=1; i*k<=mx; ++i) if(val[i*k])
tmp = __gcd(tmp, i), sum += val[i*k];
tot[k] = sum;
if(tmp ^ 1) { ans[k][1] = ans[k][0] = -1; continue; }
ans[k][1] = sum; if(val[k]) ans[k][0] = 1;
}
for(int k=mx; k>=1; --k){
for(int i=1; i<=7; ++i){
f[i][k] = C[tot[k]][i];
for(int j=k*2; j<=mx; j+=k) f[i][k] = add({f[i][k], M-f[i][j]});
if(f[i][k] && ans[k][0] == 114514) ans[k][0] = i;
}
}
for(int i=1; i<=m; ++i) cout<<ans[i][0]<<' '<<ans[i][1]<<'\n';
return 0;
}
T3 狗卡
来不及写了，先把代码挂上。
#include<bits/stdc++.h>
using namespace std;
constexpr int B = 1 << 15;
char buf[B], *p1 = buf, *p2 = buf;
#define gt() (p1==p2 && (p2=(p1=buf)+fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
template <typename T> inline void rd(T &x){
x = 0; int f = 0; char ch = gt();
for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
x = f ? -x : x;
}
template <typename T, typename ...TT> inline void rd(T &x, TT &...y){ rd(x), rd(y...); }
constexpr int N = 1.2e6 + 5;
struct Seg{ int id, l, r, len; long long sum; };
vector<int> a[N];
vector<Seg> vec;
int main(){
freopen("dog.in", "r", stdin); freopen("dog.out", "w", stdout);
int n; long long m; rd(n ,m); for(int i=1, k; i<=n; ++i){
rd(k); for(int j=1, x; j<=k; ++j)
rd(x), a[i].emplace_back(x);
vec.emplace_back(Seg{i, 0, 0, 1, a[i][0]});
for(int j=1; j<k; ++j){
Seg tmp = vec.back();
if((long long)a[i][j] * tmp.len <= tmp.sum){
vec.pop_back(); tmp.r = j, ++tmp.len, tmp.sum += a[i][j];
while(!vec.empty() && vec.back().id == i && vec.back().sum * tmp.len > tmp.sum * vec.back().len){
tmp.l = vec.back().l, tmp.len += vec.back().len, tmp.sum += vec.back().sum;
vec.pop_back();
} vec.emplace_back(tmp);
} else vec.emplace_back(Seg{i, j, j, 1, a[i][j]});
}
}
sort(vec.begin(), vec.end(), [](Seg x, Seg y){
return x.sum * y.len < y.sum * x.len;
});
long long sum = 0, ans = 0;
for(auto it : vec){
for(int i=it.l; i<=it.r; ++i){
ans += m - a[it.id][i] - sum;
sum += a[it.id][i];
}
} return printf("%lld", ans), 0;
}
T4 怪盗德基
不太会，先把 std 挂上去
#include<bits/stdc++.h>
using namespace std;
int n, a[11][11], len[11], pos[11][11], f[1024][11][11][11][11];
bitset<11> ok[11][11];
inline int dfs(int s, int l, int x, int r, int y){
if(!x && y == len[r] && s == (1 << n) - 1) return 0;
if(f[s][l][x][r][y] != -1) return f[s][l][x][r][y];
int res = 114;
if(l < n && !x){
for(int i=0; i<n; ++i) if(!(1 & (s>>i)))
for(int j=0; j<len[i]; ++j) if(ok[i][j][l])
res = min(res, dfs(s|(1<<i), i, j, r, y));
res = min(res, dfs(s, n, 0, r, y));
}
for(int i=0; i<n; ++i) if(!(1 & (s>>i)) && ok[r][y][i])
res = min(res, dfs(s|(1<<i), l, x, i, 0));
for(int i=1; i<=9; ++i){
bool flag1 = x && a[l][x-1] == i, flag2 = y < len[r] && a[r][y] == i;
if(flag1 && pos[r][i] <= y) res = min(res, dfs(s, l, x-1, r, y) + 1);
if(flag2 && pos[l][i] <= x) res = min(res, dfs(s, l, x, r, y+1) + 1);
if(flag1 && flag2) res = min(res, dfs(s, l, x-1, r, y+1) + 1);
}
return f[s][l][x][r][y] = res;
}
int main(){
freopen("hidden.in", "r", stdin); freopen("hidden.out", "w", stdout);
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
memset(pos, 0x3f, sizeof(pos));
cin>>n; for(int i=0, x, j=0; i<n; len[i]=j, j=0, ++i)
while(cin>>x && x) a[i][pos[i][x] = j++] = x;
for(int i=1; i<=9; ++i) pos[n][i] = 0;
auto check = [&](int i, int j, int z){
for(int k=0; k<len[z]; ++k){
if(j < len[i] && pos[i][a[z][k]] == j) ++j;
else if(pos[i][a[z][k]] > j) return false;
} return j == len[i];
};
for(int i=0; i<n; ++i) for(int j=0; j<len[i]; ++j) for(int z=0; z<n; ++z)
ok[i][j][z] = check(i, j, z);
for(int i=0; i<n; ++i) ok[i][0][n] = ok[n][0][i] = true;
memset(f, 0xff, sizeof(f)); int ans = 114;
for(int i=0; i<n; ++i) ans = min(ans, dfs(1<<i, i, len[i], n, 0));
return cout<<(ans >= 114 ? -1 : ans), 0;
}