NCPC 2016：简单题解

A .Artwork

pro：给定N*M的白色格子，然后Q次黑棒，输出每次加黑棒后白色连通块的数量。(N,M<1e3, Q<1e4)

sol：倒着离线做，并查集即可。

（在线做法：https://www.cnblogs.com/asdfsag/p/10485607.html

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn = 1e6 + ;
int a[][], fa[maxn], ID[][];
struct node
{
    int x1, y1, x2, y2;
    node(){}
    node(int x1, int y1, int x2, int y2):x1(x1), y1(y1), x2(x2), y2(y2){}
}c[maxn];
int Find(int x){return x == fa[x] ? x : fa[x] = Find(fa[x]);}
int dir[][] = {,,,,-,,,-};
stack<int>s;
int main()
{
    int n, m, q, tot = , x1, y1, x2, y2;
    scanf("%d%d%d", &n, &m, &q);
    for(int i = ; i <= n; i++)for(int j = ; j <= m; j++){ID[i][j] = ++tot; fa[tot] = tot;}
    for(int i = ; i <= q; i++)
    {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        c[i] = node(x1, y1, x2, y2);
        for(int x = x1; x <= x2; x++)for(int y = y1; y <= y2; y++)a[x][y]++;
    }
    int white = , cnt = ;
    for(int i = ; i <= n; i++)for(int j = ; j <= m; j++)if(!a[i][j])
    {
        white++;
        for(int k = ; k < ; k++)
        {
            int x = i + dir[k][], y = j + dir[k][];
            if(x >=  && x <= n && y >=  && y <= m && !a[x][y])
            {
                int u = ID[i][j], v = ID[x][y];
                u = Find(u);v = Find(v);
                if(u != v)fa[u] = v, cnt++;
            }
        }
    }
    s.push(white - cnt);
    for(int i = q; i >= ; i--)
    {
        for(int x = c[i].x1; x <= c[i].x2; x++)
            for(int y = c[i].y1; y <= c[i].y2; y++)
        {
            a[x][y]--;
            if(!a[x][y])
            {
                white++;
                for(int k = ; k < ; k++)
                {
                    int xx = x + dir[k][], yy = y + dir[k][];
                    if(xx >=  && xx <= n && yy >=  && yy <= m && !a[xx][yy])
                    {
                        int u = ID[x][y], v = ID[xx][yy];
                        u = Find(u);v = Find(v);
                        if(u != v)fa[u] = v, cnt++;
                    }
                }
            }
        }
        s.push(white - cnt);
    }
    while(!s.empty()){cout<<s.top()<<endl;s.pop();}
    return ;
}

C .Card Hand Sorting

pro：给定N张扑克牌，保证来自一副牌，有四种花色。现在让你重排，同种花色放一起，内部递增或者递减。 重排的方式是抽出一张牌，插入到某位置。

sol：四种花色，枚举花色的排列，再枚举每种花色内是递增还是递减。  对于每种方案，ans=N-LIS。

（写个结构体还是蛮方便的。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
map<char, int>level_num, level_color;
struct node
{
    char num, color;
    node(){}
    node(char num, char color):num(num), color(color){}
    bool operator <(const node& now)const
    {
        if(level_color[color] == level_color[now.color])//花色相同
        {
            //奇数递增，偶数递减
            if(level_color[color] & )return level_num[num] < level_num[now.num];
            else return level_num[num] > level_num[now.num];
        }
        return level_color[color] < level_color[now.color];
    }
    bool operator == (const node& now)const{return num == now.num && color == now.color;}
}a[], b[];
int dp[][], n;
int solve()
{
    memset(dp, , sizeof(dp));
    for(int i = ; i <= n; i++)
    {
        for(int j = ; j <= n; j++)
        {
            dp[i][j] = max(dp[i - ][j], dp[i][j - ]);
            if(a[i] == b[j])dp[i][j] = max(dp[i][j], dp[i - ][j - ] + );
        }
    }
    return n - dp[n][n];
}
char s[] = "shdc";
int c[];
int main()
{
    int tot = ;
    for(char i = ''; i <= ''; i++)level_num[i] = ++tot;
    level_num['T'] = ++tot;level_num['J'] = ++tot;level_num['Q'] = ++tot;
    level_num['K'] = ++tot;level_num['A'] = ++tot;
    cin >> n;
    for(int i = ; i <= n; i++){cin >> a[i].num >> a[i].color;b[i] = a[i];}
    for(int i = ; i <= ; i++)c[i] = i;
    int ans = n;
    do
    {
        for(int i = ; i <= ; i++)level_color[s[i - ]] = c[i];
        for(int i = ; i < ( << ); i++)
        {
            for(int j = ; j < ; j++)
            {
                level_color[s[j]] *= ;
                if(i & ( << j))level_color[s[j]]++;
            }
            sort(b + , b +  + n);
            ans = min(ans, solve());
        }
    }while(next_permutation(c + , c + ));
    cout<<ans<<endl;
    return ;
}

D .Daydreaming Stockbroker

pro：开始你有100元，给出N天的股价单价，你每天可以选择用当时的价格交易。限制持有股份不超过100000， 问最后有多少钱。N<365;

sol：枚举之前哪天买即可。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
ll a[maxn],dp[maxn],ans=;
int main()
{
    int N;
    scanf("%d",&N);
    rep(i,,N) scanf("%lld",&a[i]);
    dp[]=dp[]=;
    rep(i,,N){
        rep(j,,i-){
           ll t=min(dp[j]/a[j],100000LL);
           dp[i]=max(t*a[i]+dp[j]-t*a[j],dp[i]);
        }
        ans=max(ans,dp[i]);
        dp[i+]=max(dp[i+],dp[i]);
    }
    printf("%lld\n",ans);
    return ;
}

E .Exponial

pro：给定N，M。求。N^((N-1)^(N-2)...)%M；(N,M<1e9)

sol：显然欧拉降幂。 注意幂>mod时才能降幂，4以内的小于1e9，所以特判。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
int n,m;
int phi(int x){
    int ans=x;
    for(int i=;i*i<=x;i++)
        if(x%i==){
            ans/=i;ans*=i-;
            while(x%i==) x/=i;
    }
    //cout<<x<<" "<<ans<<" "<<endl;;
    if(x>) ans/=x,ans*=x-;
    return ans;
}
int ksm(int x,int y,int p){
    int ans=;
    for(;y;y>>=){
        if(y&)ans=(ll)ans*x%p;
        x=(ll)x*x%p;
    }
    return ans;
}
int dfs(int x,int y){
    if(y==) return ;
    if(x==) return ;
    if(x==) return ksm(x,,y);
    if(x==) return ksm(x,,y);
    if(x==) return ksm(x,,y);
    if(x==) return x%y;

    int ph=phi(y);
    return ksm(x,dfs(x-,ph)+ph,y);
}
int main(){
    scanf("%d%d",&n,&m);
    printf("%d",dfs(n,m));
    return ;
}

G .Game Rank

排位模拟题。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
int n,lev=,st,wins;
char s[];
int main(){
    scanf("%s",s+);
    n=strlen(s+);
    for(int i=;i<=n;i++){
        if(s[i]=='W'){
            if(lev>=&&wins>=)st+=,wins++;
            else st++,wins++;
            while(lev>&&st>)lev--,st-=;
            while(lev>&&st>)lev--,st-=;
            while(lev>&&st>)lev--,st-=;
            while(st>)lev--,st-=;
        }
        else{
            wins=;
            if(lev<=)st--;
            if(st<){
                lev++;
                if(lev>)lev=,st=;
                else if(lev>)st=;
                else if(lev>)st=;
                else st=;
            }
        }
        if(lev<=)return *puts("Legend");
    }
    printf("%d",lev);
    return ;
}

H .Highest Tower

pro：BZOJ4886

小Q正在玩一个叠塔的游戏，游戏的目标是叠出尽可能高的塔。在游戏中，一共有n张矩形卡片，其中第i张卡片的

长度为a_i，宽度为b_i。小Q需要把所有卡片按一定顺序叠成一座塔，要求对于任意一个矩形，它的长度要严格大

于它上边的任意一个矩形的长度。塔的高度为所有矩形的宽度之和。在游戏中，小Q可以将卡片翻转90度来使用，

而且必须用上全部n张卡片。请写一个程序，帮助计算小Q能叠出最高的塔的高度。

sol：定向问题。 不难证明最后是一棵或者多棵树或者环套树。 证明https://blog.csdn.net/V5ZSQ/article/details/79337446?utm_source=blogxgwz8

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
map<int,int>mp;
int ind[maxn],tot,mx[maxn],fa[maxn],tag[maxn],val[maxn]; ll ans;
int find(int x){
    if(x==fa[x]) return x;
    return fa[x]=find(fa[x]);
}
int u[maxn],v[maxn];
int main()
{
    int N,x,y;
    scanf("%d",&N);
    rep(i,,N){
        scanf("%d%d",&u[i],&v[i]);
        x=u[i]; y=v[i];
        if(!mp[x]) mp[x]=++tot,val[tot]=mx[tot]=x;
        if(!mp[y]) mp[y]=++tot,val[tot]=mx[tot]=y;
        x=u[i]=mp[x]; y=v[i]=mp[y];
        ind[x]++; ind[y]++;
    }
    rep(i,,tot) fa[i]=i;
    rep(i,,tot){
        x=find(u[i]); y=find(v[i]);
        if(tag[x]&&tag[y]) continue;
        if(x==y) tag[y]=;
        else fa[x]=y,tag[y]|=tag[x],mx[y]=max(mx[x],mx[y]);
    }
    rep(i,,tot){
        ans+=1LL*(ind[i]-)*val[i];
        if(find(i)==i&&!tag[i]) ans+=mx[i];
    }
    printf("%lld\n",ans);
    return ;
}

J .Jumbled Compass

模拟。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=;i<=;i++){
        if((n+i)%==m)return *printf("%d",i);
        if((n-i+)%==m)return *printf("%d",-i);
    }
}

K .Keeping the Dogs Apart

题意: 给定两支猫的行走路线, 都是直线行走, 求他们都在走的最近距离.

思路: 模拟即可. 每次得到min{到转折点的时间},  然后得到结束位置, 至于这个过程的最近距离,我们可以假设一个不动,那么就是点到线段的距离. 继续模拟.

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
double x[maxn][],y[maxn][],ans;
const double inf=1e200;
const double pi=*atan(1.0);
struct point{
    double x,y;
    point(double a=,double b=):x(a),y(b){}
};
int dcmp(double x){ return fabs(x)<0.0000000000001?:(x<?-:);}
point operator +(point A,point B) { return point(A.x+B.x,A.y+B.y);}
point operator -(point A,point B) { return point(A.x-B.x,A.y-B.y);}
point operator *(point A,double p){ return point(A.x*p,A.y*p);}
point operator /(point A,double p){ return point(A.x/p,A.y/p);}
point rotate(point A,double rad){
    return point(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
bool operator ==(const point& a,const point& b) {
     return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
}
double dot(point A,point B){ return A.x*B.x+A.y*B.y;}
double det(point A,point B){ return A.x*B.y-A.y*B.x;}
double dot(point O,point A,point B){ return dot(A-O,B-O);}
double det(point O,point A,point B){ return det(A-O,B-O);}
double length(point A){ return sqrt(dot(A,A));}
double angle(point A,point B){ return acos(dot(A,B)/length(A)/length(B));}
bool isPointOnSegment(point p,point a1,point a2)
{
    //点是否在线段上
    return dcmp(det(a1-p,a2-p)==&&dcmp(dot(a1-p,a2-p))<=);
}
double distoseg(point P,point A,point B)
{
    if(isPointOnSegment(P,A,B)) return 0.0;
    //点到线段距离
    if(A==B) return length(P-A);
    point v1=B-A,v2=P-A,v3=P-B;
    if(dcmp(dot(v1,v2))<) return length(v2);
    else if(dcmp(dot(v1,v3))>) return length(v3);
    return fabs(det(v1,v2)/length(v1));
}
double dist(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void solve(int A,int B,double t)
{
    double dx[],dy[];
    double d1=dist(x[A][],y[A][],x[A+][],y[A+][]);
    double d2=dist(x[B][],y[B][],x[B+][],y[B+][]);
    dx[]=(x[A+][]-x[A][])/d1; dy[]=(y[A+][]-y[A][])/d1;
    dx[]=(x[B+][]-x[B][])/d2; dy[]=(y[B+][]-y[B][])/d2;
    double sx=x[A][],sy=y[A][],tx=x[A][]+(dx[]-dx[])*t,ty=y[A][]+(dy[]-dy[])*t;
    double res=distoseg(point(x[B][],y[B][]),point(sx,sy),point(tx,ty));
    ans=min(ans,res);
    x[A][]+=dx[]*t; y[A][]+=dy[]*t;
    x[B][]+=dx[]*t; y[B][]+=dy[]*t;
}
int main()
{
    int N,M;
    scanf("%d",&N);
    rep(i,,N) scanf("%lf%lf",&x[i][],&y[i][]);
    scanf("%d",&M);
    rep(i,,M) scanf("%lf%lf",&x[i][],&y[i][]);
    ans=;
    int A=,B=;
    while(A<N&&B<M){
        double da=dist(x[A][],y[A][],x[A+][],y[A+][]);
        double db=dist(x[B][],y[B][],x[B+][],y[B+][]);
        if(da<db){
            solve(A,B,da);
            A++;
        }
        else if(da==db){
            solve(A,B,da);
            A++; B++;
        }
        else {
            solve(A,B,db);
            B++;
        }
    }
    printf("%.8lf\n",ans);
    return ;
}