UCF Local Programming Contest 2017 题解 A~I

UCF Local Programming Contest 2017 题解 A~I

A题

这题没啥好说的

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;

int main() {
    LL a,b;
    cin>>a>>b;
    LL n;
    cin>>n;
    while (n--) {
        LL c;
        cin>>c;
        cout<<c<<" ";
        LL ans=0;
        if (c>1000) {
            ans+=1000*a+(c-1000)*b;
        }
        else {
            ans=c*a;
        }
        cout<<ans<<endl;
    }
    return 0;
}

B题

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;

char g[3][9]={
    'a','b','c','d','e','f','g','h','i',
    'j','k','l','m','n','o','p','q','r',
    's','t','u','v','w','x','y','z','#',
};

int d[8][2]={1,0,-1,0,0,1,0,-1, 1,1,-1,-1,1,-1,-1,1};

void getXY(char c,int& x,int& y) {
    int n=c-'a';
    x=n/9;
    y=n%9;
}

bool isSimi(string& a,string& b) {
    bool f1=true;
    for (int i=0;i<a.length();i++) {
        if (a[i] == b[i]) continue;
        int x,y;
        getXY(a[i],x,y);
        // cout<<x<< " "<< y<<endl;
        bool f2=false;
        for (int j=0;j<8;j++) {
            int dx=d[j][0]+x, dy=d[j][1]+y;
            if (dx>=0&&dy>=0&&dx<3&&dy<9) {
                // cout<<g[dx][dy]<< " " <<endl;
                if (g[dx][dy]==b[i]) {
                    f2=true;
                    break;
                }
            }
        }
        if (!f2) return false;
    }
    return true;
}

int main() {
    string a,b;
    int n;
    cin>>n;
    while (n--) {
        cin>>a>>b;
        if (a.length()!=b.length()) {
            cout<<3<<endl;
        }
        else {
            bool flag = true;
            if (a==b) {
                cout<<1<<endl;
            }
            else {
                if (isSimi(a,b)) {
                    cout<<2<<endl;
                }
                else {
                    cout<<3<<endl;
                }
            }
        }
    }
    return 0;
}

C题

就是比较比较左，比较比较右，基础题。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;

int main() {
    LL t;
    cin>>t;
    while (t--) {
        LL n,s;
        cin>>n>>s;
        LL ans=0,c,last;
        for (int i=0;i<s;i++) {
            cin>>c;
            if (i!=0) {
                if (last == c) {
                    ans++;
                }
                else if (last < c) {
                    LL rd=c-last-1;
                    LL ld=n-c+last+1;
                    ans+=min(ld,rd);
                }
                else {
                    LL ld=last-c+1;
                    LL rd=n-last+c-1;
                    ans+=min(ld,rd);
                }
            }
            last=c;
        }
        cout<<ans<<endl;
    }
    return 0;
}

D题

这题很明显就能看出来是一道递推的题，可以用dp做，但是我们还要确定方向，所以我们直接用bfs好了，向四个方向推，数据范围也不大。

走一步之后，要么同行，要么不同行，同行判边界，异行也是判断会不会超出下一行的边界。

之后判一下是否出边界了，然后bfs本身具有最短路性质，就可以啦。

#include <bits/stdc++.h>
using namespace std;
#define mk make_pair
typedef long long LL;
const int MAXN=1e6+10;

int vis[125][85],s[125];
int T,sx,sy,ex,ey,n;
int d[4][2]={1,0,-1,0,0,1,0,-1};

struct Node
{
    int x, y, step;
    Node(int _x, int _y, int _s) : x(_x), y(_y), step(_s) {}
};

int bfs() {
    vis[sx][sy]=1;
    queue<Node> q;
    q.push(Node(sx, sy, 0));
    while (!q.empty()) {
        Node f = q.front();
        q.pop();
        if (f.x == ex && f.y == ey) {
            return f.step;
        }
        for (int i=0; i<4; i++) {
            int dx=f.x+d[i][0];
            int dy=f.y+d[i][1];
            if (dx == f.x) {
                if (dy == -1) {
                    dx--;
                    dy=s[dx];
                }
                else if (dy > s[dx]) {
                    dx++;
                    dy=0;
                }
            }
            else {
                if (dy > s[dx]) dy=s[dx];
            }
            if (dx <=0 || dy < 0 || dx > n || dy > s[dx] || vis[dx][dy]) continue;
            // cout<<dx<<" "<<dy<<endl;
            vis[dx][dy]=1;
            q.push(Node(dx,dy,f.step+1));
        }
    }
    return 0;
}

int main() {
    // freopen("in.txt","r",stdin);
    cin>>T;
    while (T--) {
        memset(vis, 0, sizeof(vis));
        cin>>n;
        for (int i=1; i<=n; i++) {
            cin>>s[i];
        }
        cin>>sx>>sy>>ex>>ey;
        cout<<bfs()<<endl;
    }
    return 0;
}

E题

这题我是用tan做的，所以要处理四个区域内的情况。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;
const double EPS=1e-5;
const double PI=acos(-1);

double dis(double x,double y) {
    return sqrt(x*x+y*y);
}

int main() {
    // freopen("in.txt","r",stdin);
    int T,K,b,d,s,w;
    cin>>T;
    while (T--) {
        cin>>w>>b>>d>>s;
        cin>>K;
        double x,y;
        int sc=0;
        while (K--) {
            cin>>x>>y;
            int flag=-1;
            if (dis(x,y) > (double)s) {
                continue;
            }
            else if (dis(x,y) > (double)d) {
                flag=1;
            }
            else if (dis(x,y) > (double)b) {
                flag=2;
            }
            else {
                flag=3;
                sc+=50;
                continue;
            }

            double k=abs(y/x);
            double ang=atan(k)/PI*180;
            double avgAng=1.0*360/w;

            if (x<0 && y>0) {
                ang=180-ang;
            }
            else if (x<0 && y<0) {
                ang=ang+180;
            }
            else if (x>0 && y<0) {
                ang=180+(180-ang);
            }
            double sum=0;
            int ts=1;
            for (int i=1;i<=w;i++) {
                sum+=avgAng;
                if (sum>ang) {
                    ts=i;
                    break;
                }
            }
            
            if (flag == 2) {
                sc+= 2 * ts;
            }
            else {
                sc+= ts;
            }

        }
        cout<<sc<<endl;
        
    }
    return 0;
}

F题

最短路的变形题啦，本来我是想用bfs状态表示的，但是看了题解之后还是决定用最短路写，之前用bfs写炸了一次空间，后来发现之前记录状态不对，但是还是最短路香。

对于城市，用4种状态作为点，4种状态之间全部用转换交通方式的费用连接。不同城市之间相同状态相连。

#include <bits/stdc++.h>
using namespace std;
#define mk make_pair
typedef long long LL;
const int MAXN=410*4;

LL ts[MAXN],d[MAXN],vis[MAXN];
vector<pair<LL,LL>> g[MAXN];

struct Node {
    LL id, d;
    Node(LL _i,LL _d):id(_i),d(_d) {}
    bool operator < (const Node& A) const {
        return d > A.d;
    }
};

void dijkstra(int s,int e) {
    memset(vis, 0 ,sizeof(vis));
    memset(d, 0x3f, sizeof(d));

    d[s]=0;
    priority_queue<Node> pq;
    pq.push(Node(s,0));

    while (!pq.empty())
    {
        Node f=pq.top();
        pq.pop();
        if (vis[f.id]) continue;
        vis[f.id]=1;
        for (auto i: g[f.id]) {
            LL v=i.first;
            LL c=i.second;
            if (f.d+c<d[v]&&!vis[v]) {
                d[v]=f.d+c;
                pq.push(Node(v,d[v]));
            }
        }
    }
    
}

int main() {
    // freopen("in.txt","r",stdin);

    ios::sync_with_stdio(false);
    int T,C,R,VAL;
    string U,V,MO;
    cin>>T;
    while (T--) {
        cin>>C;
        LL id=1;
        unordered_map<string,int> cache;
        while (C--) {
            cin>>U>>ts[id];
            cache[U]=id;
            id+=4;
        }
        
        for (int i=1;i<id;i+=4) {
            for (int j=i;j<i+4;j++) {
                for (int k=j+1;k<i+4;k++) {
                    g[j].push_back(mk(k,ts[i]));
                    g[k].push_back(mk(j,ts[i]));
                }
            }
        }
        
        cin>>R;
        while (R--) {
            cin>>U>>V>>MO>>VAL;
            LL mo=0;
            if (MO=="AIR") mo=0;
            else if (MO=="RAIL") mo=1;
            else if (MO=="SEA") mo=2;
            else mo=3;
            int u=cache[U]+mo;
            int v=cache[V]+mo;
            g[u].push_back(mk(v,VAL));
            g[v].push_back(mk(u,VAL));
        }
        cin>>U>>V;
        int des=cache[U];
        int s=0,e=id;
        for (int i=0;i<4;i++) g[s].push_back(mk(des+i,0));
        des=cache[V];
        for (int i=0;i<4;i++) g[des+i].push_back(mk(e,0));
        dijkstra(s,e);
        cout<<d[e]<<endl;
        for (LL i=0;i<id;i++) g[i].clear();
    }

    return 0;
}

G题

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;

double f[2505][10005], P[52];
int c[52];
int main() {
    // freopen("in.txt","r",stdin);
    int T,G; 
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&G);
        for (int i=1;i<=G;i++) {
            scanf("%d%lf",&c[i],&P[i]);
            // printf("%d %f
",c[i],P[i]);
        }
        int N;
        scanf("%d",&N);
        vector<double> p;
        p.push_back(0);
        for (int i=1;i<=G;i++) {
            for (int j=1;j<=c[i];j++) {
                p.push_back(P[i]);
            }
        }
        memset(f, 0, sizeof(f));
        f[0][0]=1;
        for (int i=1;i<p.size();i++) {
            for (int j=1;j<=N;j++) {
                f[i][j]+=f[i-1][j-1]*p[i];
                f[i-1][j]+=f[i-1][j-1]*(1-p[i]);
            }
        }
        double ans=0;
        for (int i=1;i<=N;i++) {
            ans+=f[p.size()-1][i];
        }
        printf("%.3f
",ans);
    }
    return 0;
}

H题

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e6+10;

int f[105][105][105],cnt[105][105],a[105];

int LIS(int l,int r) {
    vector<int> vc;
    for (int i=l;i<=r;i++) {
        auto it=lower_bound(vc.begin(),vc.end(),a[i]);
        if (it == vc.end()) {
            vc.push_back(a[i]);
        }
        else {
            int ind=it-vc.begin();
            vc[ind]=a[i];
        }
    }
    // cout<<"LIS"<<endl;
    // for (auto x: vc) cout<<x<<" ";
    // cout<<endl;
    return vc.size();
}

int main() {
    // freopen("in.txt","r",stdin);
    
    int T,N;
    cin>>T;
    while (T--) {
        cin>>N;
        for (int i=1;i<=N;i++) {
            cin>>a[i];
        }
        for (int i=1;i<=N;i++) {
            for (int j=i;j<=N;j++) {
                cnt[i][j]=LIS(i,j);
            }
        }
        memset(f, 0, sizeof(f));
        // for (int i=1;i<=N;i++) {
        //     f[1][i][i]=1;
        // }
        for (int i=1;i<=cnt[1][N];i++) {
            for (int len=1;len<=N;len++) {
                // cout<<"len:"<<len<<endl;
                for (int l=1;l+len-1<=N;l++) {
                    int r=l+len-1;
                    // cout<< f[i][l][r]<<endl;
                    if (cnt[l][r]>=i) {
                        f[i][l][r]=cnt[l][r];
                    }
                    for (int k=l;k<r;k++) {
                        f[i][l][r]=max(f[i][l][r],f[i][l][k]+f[i][k+1][r]);
                        // cout<<"i" << i<<" "<< f[i][l][r] << " "<< f[i][l][k]<< " "<< f[i][k+1][r]<<endl;
                    }
                }
            }
        }
        for (int i=1;i<N;i++) {
            cout<<f[i][1][N]<< " ";
        }
        cout<<f[N][1][N]<<endl;
    }
    return 0;
}

I题

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;

LL sum[MAXN],pos[MAXN];

LL lowbit(LL x) {
    return x&(-x);
}

void update(LL i,LL val) {
    while (i<MAXN) {
        sum[i]+=val;
        i+=lowbit(i);
    }
}

LL query(LL i) {
    LL res=0;
    while (i>0) {
        res+=sum[i];
        i-=lowbit(i);
    }
    return res;
}

int main() {
    // freopen("in.txt","r",stdin);

    int T;
    scanf("%d",&T);
    
    while (T--) {
        memset(sum,0,sizeof(sum));
        int N;
        scanf("%d",&N); 
        LL num;
        for (int i=1;i<=N;i++) {
            scanf("%lld",&num);
            pos[num]=i;
            update(i,num);
        }

        LL p=0,ans=0;
        for (int i=1;i<=N;i++) {
            LL c=pos[i];
            if (c>p) {
                LL lc=query(c-1)-query(p-1);
                LL rc=query(N)-lc;
                ans+=min(lc,rc);
            }
            else {
                LL rc=query(p-1)-query(c-1);
                LL lc=query(N)-rc;
                ans+=min(lc,rc);
            }
            update(c,-i);
            p=c+1;
        }
        printf("%lld
",ans);
    }
    return 0;
}