Codeforces Round #411 (Div. 1) 题解 (待续)

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本文解析了四道ACM竞赛题目，包括快速求解、贪心策略、树形结构的着色问题及复杂置换问题。通过高效算法实现最优解，并提供源代码示例。

A Find Amir

一眼题

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read();
    cout<<(n-1)/2<<endl;
    return 0;
}

B Minimum number of steps

注意a的数量不变，故倒着贪心一波

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
char s[1123456];
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);

    cin>>(s+1);
    int n=strlen(s+1);
    ll c=0;
    ll ans=0;
    ForD(i,n) {
        if (s[i]=='b') upd(c,1);
        else {
            upd(ans,c);
            upd(c,c);
        }
    }
    cout<<ans<<endl;
    return 0;
}

C. Ice cream coloring

考虑沿着树贪心的填能填的值中最小的，
可以证明是对的（不会证）。

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int n,m;
#define MAXN (312345)
int s[MAXN];
vi v[MAXN],e[MAXN];
int ans[MAXN]={};
int chk[MAXN]={};
void dfs(int x,int fa) {
    //找出这个点上没被染色的点 
    vi vu;
    for(auto it:v[x]) {
        if (ans[it]) chk[ans[it]]=1;
        else vu.pb(it);
    }
    int p=1;
    for(auto it:vu) {
        while(chk[p]) ++p;
        ans[it]=p++; 
    }

    for(auto it:v[x]) {
        chk[ans[it]]=0;
    }

    for(int v:e[x]) if (v^fa) dfs(v,x);
}
int main()
{
//  freopen("C.in","r",stdin);
    n=read(),m=read();
    For(i,n) {
        s[i]=read();
        For(j,s[i]) v[i].pb(read());
    }
    For(i,n-1) {
        int p=read(),q=read();
        e[p].pb(q); e[q].pb(p);
    }
    dfs(1,0);
    For(i,m) ans[i]=max(ans[i],1); 
    printf("%d\n",*max_element(ans+1,ans+1+m));
    For(i,m) printf("%d ",ans[i]);
    puts("");

    return 0;
}

E. The same permutation

暴搜找出
4个一组的换法
5个一组的换法
将2个4个一组交换的方法
然后用这3个方法把答案凑出来

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
vector<pair<int,int> > v;
void print(int i,int j) {
    printf("%d %d\n",i+1,j+1);
    printf("%d %d\n",i+3,j+1);
    printf("%d %d\n",i+2,j+2);
    printf("%d %d\n",i+3,j+2);
    printf("%d %d\n",i+1,j+3);
    printf("%d %d\n",i+2,j+1);
    printf("%d %d\n",i+0,j+0);
    printf("%d %d\n",i+0,j+3);
    printf("%d %d\n",i+1,j+0);
    printf("%d %d\n",i+2,j+3);
    printf("%d %d\n",i+3,j+0);
    printf("%d %d\n",i+1,j+2);
    printf("%d %d\n",i+0,j+1);
    printf("%d %d\n",i+0,j+2);
    printf("%d %d\n",i+2,j+0);
    printf("%d %d\n",i+3,j+3);
}
int main()
{
//  freopen("E.in","r",stdin);
//  freopen(".out","w",stdout);

    int n=read();
    if (n%4>1) puts("NO");
    else {
        puts("YES");
        for(int i=1;i<=n/4*4;i+=4)
            for(int j=i+4;j<=n/4*4;j+=4) {
                print(i,j);
            }
        for(int i=1;i<=n/4*4;i+=4) {
            if(n%4) printf("%d %d\n%d %d\n%d %d\n",i+2,n,i+2,i+3,i+3,n);
            else printf("%d %d\n",i+2,i+3);
            printf("%d %d\n",i,i+2);
            printf("%d %d\n",i+1,i+3);
            printf("%d %d\n",i+1,i+2);
            printf("%d %d\n",i,i+3);
            if(n%4)printf("%d %d\n%d %d\n%d %d\n",i,n,i,i+1,i+1,n);
            else printf("%d %d\n",i,i+1);
        }
    }  

    return 0;
}