【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.)

因为发现一个1,2,3,4,5,6,7....N的序列一次操做后能够变成

2,3,4,5,6,7...N,0的序列，这样N次事后，总会获得全部数-1的序列

也就是，我能够进行那么屡次，序列能够构形成\(K / N\)开始加1到N的序列，就是能够进行\(\lfloor \frac{K}{N} \rfloor N\)那么屡次了

剩下的就从头开始逆着往回加便可

#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.)

进行一次操做咱们直接把最大的扣到N如下，暴力进行几回复杂度不会太大

#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

若连通还每一个点就一个入度，那这是一个有向的基环外向树

把环上挂着的数给dp完，至关于每次对于儿子取一个未出现过的最小值同样的操做

而后环上的点要么取本身未出现过的最小值，要么取把这个最小值填上以后下一个没出现过的

也就是环上前一个点若是选择下一个点的第一选项，下一个点必须选择第二选项

把环上每一个点拆成两个点，按照这种关系连边，出现大小刚好为原来环点数的环时合法，若是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;
}