ARC079题解
C - Cat Snuke and a Voyage
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,M;
bool vis[MAXN];
vector<int> to[MAXN];
void Solve() {
read(N);read(M);
int a,b;
vis[N] = 1;
for(int i = 1 ; i <= M ; ++i) {
read(a);read(b);
if(a > b) swap(a,b);
if(b == N) vis[a] = 1;
to[a].pb(b);to[b].pb(a);
}
if(vis[1]) {puts("POSSIBLE");return;}
for(auto k : to[1]) {
if(vis[k]) {puts("POSSIBLE");return;}
}
puts("IMPOSSIBLE");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
D - Decrease (Contestant ver.)
由于发现一个1,2,3,4,5,6,7....N的序列一次操作后可以变成
2,3,4,5,6,7...N,0的序列，这样N次过后，总会得到所有数-1的序列
也就是，我可以进行那么多次，序列可以构造成\(K / N\)开始加1到N的序列，就是可以进行\(\lfloor \frac{K}{N} \rfloor N\)那么多次了
剩下的就从头开始逆着往回加即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int64 a[55],K;
void Solve() {
read(K);
N = 50;
a[1] = K / N;
for(int i = 2 ; i <= N ; ++i) {
a[i] = a[i - 1] + 1;
}
int t = K % N;
for(int i = 1 ; i <= t ; ++i) {
a[i] += N;
for(int j = 1 ; j <= N ; ++j) {
if(i != j) a[j]--;
}
}
out(N);enter;
for(int i = 1 ; i <= N ; ++i) {
out(a[i]);space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
E - Decrease (Judge ver.)
进行一次操作我们直接把最大的扣到N以下，暴力进行几次复杂度不会太大
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int64 a[55],ans;
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(a[i]);
while(1) {
int t = 1;
for(int i = 2 ; i <= N ; ++i) {
if(a[i] > a[t]) t = i;
}
if(a[t] < N) break;
int64 k = (a[t] - (N - 1) - 1) / N + 1;
a[t] -= k * N;
ans += k;
for(int i = 1 ; i <= N ; ++i) {
if(i != t) a[i] += k;
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
F - Namori Grundy
若连通还每个点就一个入度，那这是一个有向的基环外向树
把环上挂着的数给dp完，相当于每次对于儿子取一个未出现过的最小值一样的操作
然后环上的点要么取自己未出现过的最小值，要么取把这个最小值填上之后下一个没出现过的
也就是环上前一个点如果选择下一个点的第一选项，下一个点必须选择第二选项
把环上每个点拆成两个点，按照这种关系连边，出现大小恰好为原来环点数的环时合法，如果无环或有一个二倍点数的环就不合法
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,p[MAXN],deg[MAXN],val[MAXN];
int id[MAXN][2],tot,cir;
vector<int> t[MAXN],to[MAXN * 2];
vector<int> son[MAXN];
queue<int> q;
int dfn[MAXN * 2],low[MAXN * 2],sta[MAXN * 2],instack[MAXN * 2],idx,top;
bool flag = 0;
void Tarjan(int u) {
dfn[u] = low[u] = ++idx;
sta[++top] = u;instack[u] = 1;
for(auto v : to[u]) {
if(!dfn[v]) {Tarjan(v);low[u] = min(low[v],low[u]);}
else if(instack[u] == 1){
low[u] = min(low[u],dfn[v]);
}
}
if(low[u] == dfn[u]) {
int k = 0;
while(1) {
int x = sta[top--];
++k;
instack[x] = 2;
if(x == u) break;
}
if(k == cir) flag = 1;
}
}
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(p[i]);
deg[p[i]]++;
}
for(int i = 1 ; i <= N ; ++i) {
if(!deg[i]) q.push(i);
}
while(!q.empty()) {
int u = q.front();q.pop();
sort(son[u].begin(),son[u].end());
son[u].erase(unique(son[u].begin(),son[u].end()),son[u].end());
int m = 0;
while(m < son[u].size()) {
if(son[u][m] != m) break;
++m;
}
son[p[u]].pb(m);
if(!--deg[p[u]]) {
q.push(p[u]);
}
}
for(int i = 1 ; i <= N ; ++i) {
if(deg[i]) {
sort(son[i].begin(),son[i].end());
son[i].erase(unique(son[i].begin(),son[i].end()),son[i].end());
int m = 0,pos = 0;
while(1) {
if(pos >= son[i].size() || son[i][pos] != m) {
t[i].pb(m);
if(t[i].size() < 2) {++m;}
else break;
}
else {pos++;++m;}
}
id[i][0] = ++tot;id[i][1] = ++tot;
++cir;
}
}
for(int i = 1 ; i <= N ; ++i) {
if(deg[i]) {
int f = p[i];
to[id[f][0]].pb(id[i][t[f][0] == t[i][0]]);
to[id[f][1]].pb(id[i][t[f][1] == t[i][0]]);
}
}
for(int i = 1 ; i <= tot ; ++i) {
if(!dfn[i]) Tarjan(i);
}
if(flag) puts("POSSIBLE");
else puts("IMPOSSIBLE");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}