CAIP2021 初赛VP

写点小题，放松一下。。。

A.7-1 懂的都懂

暴力打表，然后回答询问即可。

#include<bits/stdc++.h>
using namespace std;

const int N = 100;
int a[N];

map<int, bool> mp;

inline void solve(){
    int n = 0, k = 0; cin >> n >> k;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i <= n; i++){
        for(int j = i + 1; j <= n; j++){
            for(int k = j + 1; k <= n; k++){
                for(int l = k + 1; l <= n; l++){
                    mp[a[i] + a[j] + a[k] + a[l]] = true;
                }
            }
        }
    }
    for(int i = 1; i <= k; i++){
        int m; cin >> m;
        bool flag = true;
        for(int i = 1; i <= m; i++){
            int x; cin >> x;
            if(!mp.count(x * 4)) flag = false;
        }
        if(flag) cout << "Yes\n";
        else cout << "No\n";
    }
}

signed main(){
    solve();
    return 0;
}

B.7-2 芬兰木棋

分象限和坐标轴按斜率排序，然后每个区域按顺序进行遍历，动态改变方向统计即可。

最大分数一定是所有木棋之和。可以assert检验答案对不对。

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e5 + 10, JD = 1e-5;


struct node{
    int x, y, p;
    double dis, k;
    const bool operator< (const node & x) const { return k < x.k || fabs(k - x.k) <= JD && dis < x.dis; }
}a[9][N];
int num[9] = {0};

inline void solve(){
    int n = 0; cin >> n;
    //int sum = 0;
    for(int i = 1; i <= n; i++){
        int x, y, p; cin >> x >> y >> p; //sum += p;
        double dis = pow((double)(x * x  + y * y), 0.5);
        double k = 1.0 * y / (1.0 * x);
        if(x > 0 && y > 0) a[1][++num[1]] = {x, y, p, dis, k};
        if(x < 0 && y > 0) a[2][++num[2]] = {x, y, p, dis, k};
        if(x < 0 && y < 0) a[3][++num[3]] = {x, y, p, dis, k};
        if(x > 0 && y < 0) a[4][++num[4]] = {x, y, p, dis, k};
        if(x > 0 && y == 0) a[5][++num[5]] = {x, y, p, dis, 0};
        if(x < 0 && y == 0) a[6][++num[6]] = {x, y, p, dis, 0};
        if(x == 0 && y > 0) a[7][++num[7]] = {x, y, p, dis, 0};
        if(x == 0 && y < 0) a[8][++num[8]] = {x, y, p, dis, 0};
    }
    for(int i = 1; i <= 8; i++) sort(a[i] + 1, a[i] + 1 + num[i]);

    int cnt = 0, ans = 0, ansop = 0;
    for(int x = 1; x <= 8; x++){
        for(int i = 1; i <= num[x]; i++){
            if(fabs(a[x][i].k - a[x][i - 1].k) <= JD){
                if(a[x][i].p == 1) cnt++;
                else{
                    if(cnt) ansop++, ans += cnt, cnt = 0;
                    ansop++, ans += a[x][i].p;
                }
            }
            else{
                if(cnt) ansop++, ans += cnt, cnt = 0;
                if(a[x][i].p == 1) cnt++;
                else ansop++, ans += a[x][i].p;
            }
        }
        if(cnt) ansop++, ans += cnt, cnt = 0;
    }
    //assert(sum == ans);
    cout << ans << ' ' << ansop << endl;
}

signed main(){
    ios_base::sync_with_stdio(false);
    solve();
    return 0;
}

C.7-3 打怪升级

先跑Floyd再跑Dijkstra，代码没存…

属于是什么算法都能杂一起了。。。

D.7-4 疫情防控

典中典之并查集维护倒序加点

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;

const int N = 1e5 + 10;
vector<int> g[N];

struct unionfind{
    int par[N];
    unionfind(int n) { for(int i = 1; i <= n; i++) par[i] = i; }
    int find(int u){ return (u == par[u] ? u : par[u] = find(par[u])); }
    void merge(int u, int v){
        int fau = find(u), fav = find(v);
        if(fau != fav) par[fav] = u;
    }  
};

bitset<N> vis;

using pii = pair<int, int>;


struct query{
    int c, q; 
    vector<pii> qu;
}a[N];

int ans[N];

inline void solve(){
    int n, m, d; cin >> n >> m >> d;
    for(int i = 1; i <= m; i++){
        int u, v; cin >> u >> v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
    }
    vis.reset();
    for(int i = 1; i <= d; i++){
        int c, q; cin >> c >> q;
        vis[c] = true;
        a[i] = {c, q};
        for(int j = 1; j <= q; j++){
            int u, v; cin >> u >> v;
            a[i].qu.emplace_back(pii{u, v});
        }
    }
    unionfind uf(n);
    for(int i = 1; i <= n; i++){
        if(!vis[i]){
            for(auto v : g[i]) 
                if(!vis[v]) uf.merge(i, v);
        } 
    }
    for(int i = d; i >= 1; i--){
        int cnt = 0;
        for(auto q : a[i].qu){
            if(uf.find(q.first) != uf.find(q.second)) cnt++; 
        }
        int u = a[i].c;
        vis[u] = false;
        for(auto v : g[u]){
            if(!vis[v]) uf.merge(u, v);
        }
        ans[i] = cnt;
    }
    for(int i = 1; i <= d; i++) cout << ans[i] << endl;
}

signed main(){
    solve();
    return 0;
}