PAT2021年冬季考试甲级，摸鱼游记92分

Posted 2022-01-19 小哈里

T1，简单模拟，20/20分

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int main()
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, m;  cin>>n>>m;
    for(int i = 1; i <= m; i++)
        vector<int>vc;
        map<int,int>ma;
        for(int j = 1; j <= n; j++)
            int x;  cin>>x;
            vc.push_back(x);
            ma[x]++;
        
        int mx = 0;
        for(int j = 0; j < n; j++)
            vc[j] = n-ma[vc[j]];
            mx = max(mx, vc[j]);
        
        for(int j = 0; j < n; j++)
            if(vc[j]==mx)
                a[j+1]++;
            
        
    
    int mx = 0;  vector<int>res;
    for(int i = 1; i <= n; i++)
        if(a[i]>mx)
            mx = a[i];
            res.clear();
            res.push_back(i);
        else if(a[i]==mx)
            res.push_back(i);
        
    
    cout<<res[0];
    // for(int i = 0; i < res.size(); i++)
    //     cout<<res[i]<<" ";
    // 
    return 0;

T2，模拟链表，25/25分

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+10;
int a[maxn], ans[maxn];
int nxt[maxn], in[maxn], pre[maxn];
int main()
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    memset(pre,-1,sizeof(pre));
    int n;  cin>>n;
    int ed;
    for(int i = 0; i < n; i++)
        int x; cin>>x;
        a[i]= x;
        nxt[i] = x;
        if(x!=-1)pre[x] = i;
        else
            ed = i;
        
    
    vector<int>vc;
    for(int x = ed; x!=-1; x = pre[x])
        vc.push_back(x);
    
    reverse(vc.begin(),vc.end());
    for(int i = 0; i < n; i++)
        ans[vc[i]] = i;
    
    for(int i = 0; i < n ; i++)
        if(i!=0)cout<<" ";
        cout<<ans[i]+1;
    
    return 0;

T3，简单树遍历，25/25分

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
vector<int>G[maxn];
int in[maxn], ans[maxn];
int dfs(int x)
    int res = 0;
    for(int to : G[x])
        res += dfs(to);
    
    return ans[x] = res+G[x].size();

int main()
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n;  cin>>n;
    for(int i = 2; i <= n; i++)
        int x;  cin>>x;
        G[x].push_back(i);
        in[i]++;
    
    int rt;
    for(int i = 1; i <= n; i++)
        if(in[i]==0)
            rt = i; break;
        
    
    dfs(rt);
    int m;  cin>>m;
    for(int i = 0; i< m; i++)
        int x;  cin>>x;
        cout<<ans[x]+1<<"\\n";
    
    return 0;

T4，稍复杂模拟，22/30分

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
const int inf = 1e9+10;

struct tim int hh, mm; ;
void add(tim &t, int min)
    t.mm += min;
    while(t.mm>=60)
        t.mm -= 60;
        t.hh++;
    

int smal(tim x, tim y)
    if(x.hh!=y.hh)return x.hh<y.hh;
    return x.mm<y.mm;


struct node tim t; int Y, P; a[1010];
int e[2020][2020];

int main()
    //input
    int n, m;  scanf("%d%d", &n, &m);
    tim st;  scanf("%d:%d", &st.hh,&st.mm);
    a[0].t = st;
    a[0].P = a[0].Y = 0;
    for(int i = 1; i <= n; i++)
        scanf("%d:%d %d %d",&a[i].t.hh,&a[i].t.mm, &a[i].Y, &a[i].P);
    
    //memset(e,0x3f,sizeof(e));
    for(int i = 0; i <= n; i++)
        for(int j=  0; j <= n; j++)
            if(i!=j)e[i][j] = inf;
            else e[i][j] = 0;
        
    
    for(int i = 1; i <= m; i++)
        int u, v, w;  scanf("%d%d%d",&u, &v, &w);
        e[u][v] = min(e[u][v], w);
        e[v][u] = min(e[v][u], w);
        if(w>120) while(1);
    
    //min_dis,floyd
    for(int k = 0; k <= n; k++)
        for(int i = 0; i <= n; i++)
            for(int j = 0; j <= n; j++)
                if(i!=j)e[i][j] = min(e[i][j], e[i][k]+e[k][j]);
            
        
    
    // for(int i = 0; i <= n; i++)
    //     for(int j = 0; j <= n; j++)
    //         cout<<e[i][j]<<" ";
    //     
    //     cout<<"\\n";
    // 

    //跑K次
    int K;  cin>>K;
    tim at= tim23,59;  int ap = -1e9+10;//肯定比23:59早，亏钱亏钱
    while(K--)
        tim tt = st;  int tp=0; int now = 0;//0元钱，在起点，时间st
        int ok = 1;  set<int>se;

        //printf("%d: %d %02d:%02d\\n", K+1, tp, tt.hh,tt.mm);
        for(int i = 1; i <= n+1; i++)
            se.insert(now);
            int x;
            if(i!=n+1)cin>>x; else x = 0;//下一个要去的点
            if(e[now][x]==inf)ok = 0;//无法到达就再见
            if(x>n||x<0)ok = 0;
            //if(a[x].bad==1)ok = 0;//垃圾点再见
            if(ok==0)continue;

            add(tt, e[now][x]);//用时间到达
            tp += a[x].Y;//获得那么多钱
            //小于等于deadline
            if(smal(tt,(a[x].t))==1 )//  || (tt.hh==a[x].t.hh&&tt.mm==a[x].t.mm
                now = x; continue;
            else
                //大于deadline，扣钱
                tp -= a[x].P;
            
            now = x;
            //printf("%d: %d %02d:%02d\\n", K+1, tp, tt.hh,tt.mm);
        
        if(se.size()!=n+1)continue;//如果有点没有到达，那么不算!
        //if(ok==0)continue;//如果无法到达点，那么慢不算

        //printf("%d: %d %02d:%02d\\n", K+1, tp, tt.hh,tt.mm);
        if(tp>ap)//赚的钱更多优先
            ap = tp;  at = tt;
        else if(tp==ap)// 钱一样，时间短优先
            if(smal(tt,at)==1)
                at = tt;
            
        
    
    printf("%d %02d:%02d\\n", ap, at.hh,at.mm);
    return 0;