【AtCoder】ARC079

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;
}