Codeforces Round #394 (Div. 2) 题解

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题目连接：http://codeforces.com/contest/761

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A. Dasha and Stairs

解题思路：

判断一下两者相减的绝度值是否为小于2。注意特判两个0的情况

时间复杂度：O(1)
空间复杂度：O(1)

代码如下：

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#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m;int main(){    cin >> n>>m;    if(n==0&&m==0)    {        cout << "NO"<<endl;        return 0;    }    if(abs(n-m)<=1)    {        cout << "YES"<<endl;    }    else    {        cout << "NO"<<endl;    }    return 0;   } 

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B - Dasha and friends

题目大意

给两个序列描述圆上的点，判断这两个圆是否可以视为同一个圆。

解题思路：

暴力搜索，两个序列元素分别一一对应，直到找到所有点距离都相同的情况。注意环形特征- ->取余

时间复杂度：O(n^2)
空间复杂度：O(n)

代码如下：

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#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int judge(int x);int n,l;vi a,b;int main(){    int t;    cin >> n>>l;    rep(i,1,n)    {        cin >> t;        a.push_back(t);    }     rep(i,1,n)    {        cin >> t;        b.push_back(t);    }     for(int i=0;i<n;i++)    {        if(judge(i))        {            cout <<"YES"<<endl;            return 0;        }    }    cout <<"NO"<<endl;    return 0;   } int judge(int x){    int xx ;    if(a[0]<b[(x)%n])    {        xx = a[0]+l-b[(x)%n];    }     else    {        xx = a[0]-b[(x)%n];    }    //cout <<xx<<endl;    for(int i=1;i<n;i++)    {        int yy ;        if(a[i]<b[(i+x)%n])        {            yy = a[i]+l-b[(i+x)%n];        }        else        {            yy =a[i]- b[(i+x)%n];        }        if(yy!=xx)        {        //  cout << i<<" "<<x<<endl;            return 0;        }    }    return 1;} 

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C. Dasha and Password

题目大意

在多个字符串中组成一个符合条件的密码，问最小移动次数

解题思路：

对三种情况暴力搜索，分别找到最小，注意判断三者最小是否在一个字符串上。

时间复杂度：O(n^2)
空间复杂度：O(n)

代码如下：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int n,m;vector<string> a;int ffb=-1,ffc=-1,ffd=-1;string s;int b,c,d,b1,b2,c1,c2,d1,d2;int main(){    cin >>n>>m;    rep(i,1,n)    {        cin >> s;        if(s[0]=='#'||s[0]=='*'||s[0]=='&')        {            ffd = i-1;            d++;//计数        }        else if(s[0]>='0'&&s[0]<='9')        {            ffc =i-1;             c++;//计数        }        else         {            ffb = i-1;            b++;        }        a.push_back(s);    }    int ct =0;    //cout << b<<" "<<c<<" "<<d<<endl;    //如果只有一个字符串满足条件，，删去    if(b==1)    {        a.erase(a.begin()+ffb);        n--;    }    if(c==1)    {        a.erase(a.begin()+ffc);        n--;    }    if(d==1)    {        a.erase(a.begin()+ffd);        n--;    }    //遍历找最小移动次数，找到后删除，防止重复    if(b==0)    {        ct++;        int tt= inf_int;        for(int i=0;i<n;i++)        {            int t=0;            //s.clear();            s = a[i];            for(int j1=0,j2=0;j1!=j2||t==0;)            {                t++;                j1 = (j1-1+m)%m;                j2 = (j2+1)%m;                if(s[j1]>='a'&&s[j1]<='z'                ||s[j2]>='a'&&s[j2]<='z')                {                    if(tt>t)                    {                        b1 = i;                        tt= t;                        break;                    }                }            }        //  cout <<tt<<endl;        }        if(tt==inf_int)            tt=0;        b2 = tt;        if(b2!=0)        {            a.erase(a.begin()+b1);            n--;        }    }    if(c==0)    {        ct++;        int tt= inf_int;        for(int i=0;i<n;i++)        {            int t=0;            s = a[i];            for(int j1=0,j2=0;j1!=j2||t==0;)            {                t++;                j1 = (j1-1+m)%m;                j2 = (j2+1)%m;                if(s[j1]>='0'&&s[j1]<='9'                ||s[j2]>='0'&&s[j2]<='9')                {                    if(tt>t)                    {                        c1 = i;                        tt= t;                        break;                    }                }            }        }        if(tt==inf_int)            tt=0;        c2=tt;        if(c2!=0)        {            a.erase(a.begin()+c1);            n--;        }    }    if(d==0)    {        ct++;        int tt= inf_int;        for(int i=0;i<n;i++)        {            int t=0;            s = a[i];            for(int j1=0,j2=0;j1!=j2||t==0;)            {                t++;                j1 = (j1-1+m)%m;                j2 = (j2+1)%m;                if(s[j1]=='#'||s[j1]=='*'||s[j1]=='&'                ||s[j2]=='#'||s[j2]=='*'||s[j2]=='&')                {                    if(tt>t)                    {                        d1 = i;                        tt= t;                        break;                    }                }            }        }        if(tt==inf_int)            tt=0;        d2=tt;        if(d2!=0)        {            a.erase(a.begin()+d1);            n--;        }    }    if(ct==0)    {        cout <<0<<endl;        return 0;       }     else if(ct==1)    {        cout << b2+c2+d2<<endl;        return 0;    }    else    {        if(b)        {                cout << c2+d2<<endl;                return 0;        }        else if(c)        {                cout << b2+d2<<endl;                return 0;        }        else        {                cout << b2+c2<<endl;                return 0;        }    }    return 0;   } 

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D. Dasha and Very Difficult Problem

题目大意

p数组代表数组c的大小排列，而且c中元素互不相同，c[i] = b[i]-a[i],
a,b,元素大小限制为 l ≤ ai ，bi≤ r
给定数组a，求满足条件的数组b

解题思路：

贪心，首先找出c中最大的元素，然后给相应的b赋值为r，然后按照元素大小，按照每次c减小1的做法求出b，如果b大于r ，则取b为r，如果b小于l，则不符合条件，输出-1 。

时间复杂度：O(n*log(n))
空间复杂度：O(n)

代码如下：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;const int N = 1e5+5; typedef struct node{    int x;    int index;}NODE;bool cmp(NODE x,NODE y){    return x.x>y.x;}int n,l,r;int a[N];NODE p[N];int b[N];set<int> se;int main(){    cin >>n>>l>>r;    rep(i,1,n)    {        cin >> a[i];        }     rep(i,1,n)    {        cin >> p[i].x;        p[i].index = i;     }     sort(p+1,p+n+1,cmp);    b[p[1].index] = r;    rep(i,2,n)    {        int tt = a[p[i].index]+b[p[i-1].index]-a[p[i-1].index]-1;        if(tt>r)        {            b[p[i].index] = r;        }        else if(tt>=l)        {            b[p[i].index] = tt;        }        else        {            cout <<-1<<endl;            return 0;        }    }    rep(i,1,n)    {        cout << b[i]<<" ";    }    cout <<endl;    return 0;   } 

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E. Dasha and Puzzle(构造题+DFS)

题目大意

给定一颗树，判断其在笛卡尔平面内，是否有边与坐标平行，且不相交的情况。

解题思路：

如果有一个顶点的出度超过4，则不满足题意。
否则根据长度构造树，使每层的长度有较大差距，就保证了其不相交。
利用DFS放点。我用的是对2的幂级来确定长度。

时间复杂度：O(n)
空间复杂度：O(n)

代码如下：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{1,0},{0,-1},{-1,0},{0,1}};const ll N = 1e18; typedef struct node{    ll x;    ll y;}NODE;void dfs(int x,int chang,int dr);int flag = 0;int book1[40];int n,u,v;int in[40];NODE pos[40];int ct = 0;vi mpp[40];int main(){    cin >>n;    rep(i,1,n-1)    {        cin >> u>> v;        in[v]++;        mpp[u].push_back(v);        mpp[v].push_back(u);    }    rep(i,1,n)     {        if(mpp[i].size()>4)        {            cout <<"NO"<<endl;            return 0;        }    }    rep(i,1,n)     {        if(in[i]==0)        {            pos[i].x = N/2;            pos[i].y = N/2;            book1[i]=  1;             dfs(i,40,-100);            break;        }    }    cout <<"YES"<<endl;    rep(i,1,n)        cout << pos[i].x<<" "<<pos[i].y <<endl;     return 0;   } void dfs(int x,int chang,int dr){    ct =0;    for(int i=1;i<=n;i++)    {        ct += book1[i];    }    //cout <<ct<<endl;    if(flag) return ;    if(ct>=n)    {        flag = 1;        return ;    }    for(int i=0;i<4;i++)    {        if((dr+2)%4==i) continue;        for(int k=0;k<mpp[x].size();k++)        {            if(book1[mpp[x][k]]==1) continue;            ll xx = pos[x].x + dir[i][0]*(((ll)1<<chang)*100-3*mpp[x][k]);            ll yy = pos[x].y + dir[i][1]*(((ll)1<<chang)*100-3*mpp[x][k]);            book1[mpp[x][k]]=1;        //  cout << mpp[x][k]<<endl;            pos[mpp[x][k]].x = xx;            pos[mpp[x][k]].y = yy;            dfs(mpp[x][k],chang-1,i);            break;        }    }}