2022NOIP A层联测29
从今天开始学习写一句话题解
A
把合法的序列搜出来建 \(AC\)自动机，大于 \(max(p,q,r)\)的合并，然后就是求长度为 \(n\) ，匹配过关键点的串数量，容易解决
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
int read(){
int x = 0; char c = getchar();
while(!isdigit(c))c = getchar();
do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
return x;
}
const int maxn = 55;
const int mod = 998244353;
void add(int &x, int y){x += y; if(x >= mod)x -= mod;}
int n, m, p, q, r, mx;
int f[maxn][100005][2];
int a[maxn];
struct AC{
int son[100005][7], root = 1, cnt = 1, fail[100005];
bool flag[100005];
void insert(int len){
int now = root;
for(int i = 1; i <= len; ++i){
if(!son[now][a[i]])son[now][a[i]] = ++cnt;
now = son[now][a[i]];
}
flag[now] = true;
}
queue<int>q;
void built(){
for(int i = 0; i <= mx; ++i)
if(son[root][i])q.push(son[root][i]), fail[son[root][i]] = root;
else son[root][i] = root;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i <= mx; ++i)
if(son[x][i])q.push(son[x][i]), fail[son[x][i]] = son[fail[x]][i];
else son[x][i] = son[fail[x]][i];
}
}
int trans(){
f[0][1][0] = 1;
for(int len = 0; len < n; ++len)
for(int i = 1; i <= cnt; ++i)if(f[len][i][0] || f[len][i][1]){
for(int j = 1; j <= mx; ++j){
int v = son[i][j], fl = flag[v];
add(f[len + 1][v][fl], f[len][i][0]);
add(f[len + 1][v][1], f[len][i][1]);
}
if(m > mx){
add(f[len + 1][1][0], 1ll * f[len][i][0] * (m - mx) % mod);
add(f[len + 1][1][1], 1ll * f[len][i][1] * (m - mx) % mod);
}
}
int ans = 0;
for(int i = 1; i <= cnt; ++i)add(ans, f[n][i][1]);
return ans;
}
}A;
void dfs(int pos, int sp, int sq, int sr){
if(pos > n + 1)return;
if(sp < p)
for(int i = 1; i <= min(p - sp, m); ++i)a[pos] = i, dfs(pos + 1, sp + i, sq, sr);
else{
if(sq < q)
for(int i = 1; i <= min(q - sq, m); ++i)a[pos] = i, dfs(pos + 1, sp, sq + i, sr);
else{
if(sr < r)
for(int i = 1; i <= min(r - sr, m); ++i) a[pos] = i, dfs(pos + 1, sp, sq, sr + i);
else A.insert(pos - 1);
}
}
}
int sol(){
dfs(1, 0, 0, 0);
mx = max({p, q, r}); mx = min(mx, m);
A.built();
return A.trans();
}
int main(){
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
n = read(), m = read(), p = read(), q = read(), r = read();
printf("%d\n",sol());
return 0;
}
B
扩展欧拉定理
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1000006;
ll mod, phi;
ll qpow(ll x, ll y){
ll ans = 1;
for(; y; y >>= 1, x = x * x % mod)if(y & 1)ans = ans * x % mod;
return ans;
}
char a[maxn], b[maxn];
ll la, lb, n, k;
int main(){
freopen("b.in","r",stdin);
freopen("b.out","w",stdout);
scanf("%s%s",a + 1, b + 1);
scanf("%lld",&mod);
la = strlen(a + 1); lb = strlen(b + 1);
int m = mod; phi = mod;
for(int i = 2; i * i <= m; ++i)if((m % i) == 0){
phi = phi / i * (i - 1);
while(m % i == 0)m /= i;
}
if(m != 1){phi = phi / m * (m - 1);}
for(int i = 1; i <= la; ++i)n = (n * 10 + (a[i] ^ 48)) % phi;
bool flag = false;
if(la <= 10){
ll now = 0;
for(int i = 1; i <= la; ++i)now = (now * 10 + (a[i] ^ 48));
if(now < phi)flag = true;
}
for(int i = 1; i <= lb; ++i)k = (k * 10 + (b[i] ^ 48)) % mod;
n = (n + phi - 1) % phi;
ll ans = 1ll * (k - 1) * qpow(k, flag ? n : n + phi) % mod;
printf("%lld\n",ans);
return 0;
}
C
容易证明最优解一定是 \(u - m - v - m - w\)
（考虑反证）
然后枚举一下 \(m\) 即可
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
int read(){
int x = 0; char c = getchar();
while(!isdigit(c))c = getchar();
do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
return x;
}
const int maxn = 300005;
int dis[3][maxn];
ll val[maxn];
queue<int>q;
vector<int>g[maxn];
int n, m;
void bfs(int s, int id){
for(int i = 1; i <= n; ++i)dis[id][i] = maxn;
dis[id][s] = 0; q.push(s);
while(!q.empty()){
int x = q.front(); q.pop();
for(int v : g[x])if(dis[id][v] > dis[id][x] + 1){
dis[id][v] = dis[id][x] + 1;
q.push(v);
}
}
}
int x, y, z;
int main(){
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
n = read(), m = read(), x = read(), y = read(), z = read();
for(int i = 1; i <= m; ++i){
int u = read(), v = read(); val[i] = read();
g[u].push_back(v);
g[v].push_back(u);
}
sort(val + 1, val + m + 1);
for(int i = 1; i <= m; ++i)val[i] += val[i - 1];
bfs(x, 0); bfs(y, 1); bfs(z, 2);
ll ans = 4e18;
for(int i = 1; i <= n; ++i){
if(dis[0][i] + dis[1][i] + dis[2][i] <= m){
ll now = val[dis[1][i]] + val[dis[0][i] + dis[1][i] + dis[2][i]];
ans = min(ans, now);
}
}
printf("%lld\n",ans);
return 0;
}
D
容易发现，跟联通块个数有关，如果度数为偶数，欧拉回路即可
如果存在奇数，那么偶数显然不用管，只有把奇数消完就行
然后一个联通块提笔次数为 \(\sum [deg \& 1] / 2 - 1\) 因为每次可以消去两个奇度数
分横竖线考虑
先合并线段，然后进行排序，依次考虑每条横/竖线，按顺序处理交点
用 \(map\) 记一下点的编号映射
其实就没了
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
int read(){
int x = 0;bool f = 0; char c = getchar();
while(!isdigit(c)){f = c =='-'; c = getchar();}
do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
return f ? -x : x;
}
const int maxn = 2005;
struct DSU{
int f[maxn * maxn * 10];
int fa(int x){return x == f[x] ? x : f[x] = fa(f[x]);}
void merge(int x, int y){f[fa(y)] = fa(x);}
}S;
struct Line{
int x1, x2, y;
friend bool operator < (const Line &x, const Line &y){return x.y < y.y;}
}h[maxn], z[maxn];
int n, m, ch, cz;
bool check(Line H, Line Z){return H.x1 <= Z.y && H.x2 >= Z.y && Z.x1 <= H.y && Z.x2 >= H.y;}
bool cover(Line L, Line R){return L.y == R.y && ((L.x1 >= R.x1 && L.x1 <= R.x2) || (L.x2 >= R.x1 && L.x2 <= R.x2));}
bool del[maxn];
void uniqu(Line a[], int &len){
int nlen = 0;
for(int i = 1; i <= len; ++i)del[i] = false;
for(int i = 1; i <= len; ++i)
for(int j = 1; j < i; ++j)if(!del[j] && (cover(a[i], a[j]) || cover(a[j], a[i]))){
a[i].x1 = min(a[i].x1, a[j].x1);
a[i].x2 = max(a[i].x2, a[j].x2);
del[j] = true;
}
for(int i = 1; i <= len; ++i)if(!del[i])a[++nlen] = a[i];
len = nlen;
}
int tot, deg[maxn * maxn * 10], ans;
map<pii, int> id;
void link(int u, int v){
// printf("link : %d %d\n",u, v);
++deg[u]; ++deg[v]; S.merge(u, v);
}
void New(pii now){
// printf("new : %d %d\n",now.first, now.second);
id[now] = ++tot; S.f[tot] = tot;
}
void LINK(){
for(int i = 1; i <= ch; ++i){
pii now = pii(h[i].x1, h[i].y);
if(!id[now])New(now); int las = id[now];
for(int j = 1; j <= cz; ++j)if(check(h[i], z[j])){
pii now = pii(z[j].y, h[i].y);
if(!id[now])New(now);
if(las != id[now])link(las, id[now]);
las = id[now];
}
now = pii(h[i].x2, h[i].y);
if(!id[now])New(now);
if(id[now] != las)link(las, id[now]);
}
for(int i = 1; i <= cz; ++i){
pii now = pii(z[i].y, z[i].x1);
if(!id[now])New(now); int las = id[now];
for(int j = 1; j <= ch; ++j)if(check(h[j], z[i])){
pii now = pii(z[i].y, h[j].y);
if(!id[now])New(now);
if(las != id[now])link(las, id[now]);
las = id[now];
}
now = pii(z[i].y, z[i].x2);
if(!id[now])New(now);
if(id[now] != las)link(las, id[now]);
}
}
int val[maxn * maxn * 10];
void CUT(){
ans = -1;
for(int i = 1; i <= tot; ++i)ans += S.fa(i) == i;
if(!(m & 1))return;
for(int i = 1; i <= tot; ++i)if(deg[i] & 1)++val[S.fa(i)];
for(int i = 1; i <= tot; ++i)if(S.fa(i) == i)ans = (ans + max(val[i] / 2 - 1, 0));
// for(int i = 1; i <= tot; ++i)if(S.fa(i) == i)printf("%d ",val[i]); printf("\n");
}
int main(){
freopen("d.in","r",stdin);
freopen("d.out","w",stdout);
n = read(), m = read();
for(int i = 1; i <= n; ++i){
int a = read(), b = read(), c = read(), d = read();
if(a == c)z[++cz] = {min(b, d), max(b, d), a};
else h[++ch] = {min(a, c), max(a, c), b};
}
//for(int i = 1; i <= ch; ++i)printf("%d %d %d %d\n", h[i].x1, h[i].y, h[i].x2, h[i].y);
//for(int i = 1; i <= cz; ++i)printf("%d %d %d %d\n", z[i].y, z[i].x1, z[i].y, z[i].x2);
uniqu(z, cz); uniqu(h, ch); sort(z + 1, z + cz + 1); sort(h + 1, h + ch + 1);
//for(int i = 1; i <= ch; ++i)printf("%d %d %d %d\n", h[i].x1, h[i].y, h[i].x2, h[i].y);
//for(int i = 1; i <= cz; ++i)printf("%d %d %d %d\n", z[i].y, z[i].x1, z[i].y, z[i].x2);
LINK(); CUT();
printf("%d\n",ans);
return 0;
}
好像写了不止一句，，，，
下次主要
