SMU Winter 2024 Round #3 (Div.2)

1.New game（是一个拓扑排序板子稍微变形，但是循环加优先队列也可以做）

板子

#include <iostream>
#include <algorithm>
#include <utility>
#include <queue>
#include <cstring>
#define x first
#define y second

using namespace std;

const int N = 100010;

int m,n;
int h[N] , e[N] , ne[N] , idx; //邻接表存图
int d[N]; //入度
queue <int> q,Q;

//一个有像无环图至少有一个入度为0的点

void add(int a,int b){
    e[idx] = b , ne[idx] = h[a] , h[a] = idx++ ;
}

bool topsort(){
    for(int i=1;i<=m;i++) if(!d[i]) q.push(i); //将入度为0的点存入队列
    while(q.size()){
        int t = q.front();
        Q.push(t);
        q.pop();
        for(int i = h[t]; i != -1; i = ne[i]){
            d[e[i]]--;
            if(d[e[i]]==0) q.push(e[i]);
        }
    }
    return Q.size()==m;
}

int main(){
    memset(h , -1 , sizeof h);
    int t;
    cin>>t;
    while(t--) {
        cin >> m >> n;
        for (int i = 1; i <= n; i++) {
            int a, b;
            cin >> a >> b;
            add(a, b);
            d[b]++;
        }
        if (topsort()) while (Q.size()) cout << Q.front() << ' ', Q.pop();
        else cout << -1 << endl;
    }
    return 0;
}

循环做法

#include <iostream>
#include <algorithm>
#include <utility>
#include <queue>
#include <cstring>
using namespace std;
const int N=1e5+10;
priority_queue<int,vector<int>,greater<int>> head;
vector<int >g[N];
vector<int>ans;
int n,m;
int in[N];
void solve()
{
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        g[i].clear();
        in[i]=0;
    }
    for(int i=1;i<=m;i++)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        g[a].push_back(b);
        in[b]++;
    }
    for(int i=1;i<=n;i++)if(in[i]==0)head.push(i);
    while(head.size())
    {
        int t=head.top();
        head.pop();
        printf("%d ",t);
        for(auto val:g[t])
        {
            in[val]--;
            if(in[val]==0)head.push(val);
        }
    }
    puts("");
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}

2.Kwords Find Kth Element（现将每个序列排序，再二分为mid，查询小于等于mid的数字数量sum，如果sum大于等于k就调低上限，否则调升）

#include<bits/stdc++.h>
using namespace std;
vector<long long>a[100005];
long long aa,lll,m[100005] ,l,r,mid ,mid1,ll,rr,n,q,kk;
long long k[100005],sum ,ans, ansans;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n;
    for (int i=1;i<=n;i++) {
        cin >> m[i];
        a[i].push_back(0);
        for (int j = 1; j <= m[i]; j++) {
            cin >> aa;
            a[i].push_back(aa);
        }
        sort(a[i].begin() + 1, a[i].end());
    }
    cin>>q;
    for (int i=1;i<=q;i++)
    {
        cin>>lll;
        for (int j=1;j<=lll;j++) {
            cin >> k[j];
        }
        cin>>kk;
        l=1;r=1e9;
        while (l<=r)
        {
            mid=(l+r)/2;

            sum=0;

            for (int j=1;j<=lll;j++) {
                ll = 1;
                rr = m[k[j]];
                ans = 0;
                while (ll <= rr) {
                    mid1 = (ll + rr) / 2;
                    if (a[k[j]][mid1] <= mid) {
                        ans = mid1;
                        ll = mid1 + 1;
                    } else rr = mid1 - 1;

                }

                sum = sum + ans;
            }
            if(sum>=kk)
            {
                ansans=mid;
                r=mid-1;
            }
            else
                l=mid+1;
        }
        cout<<ansans<<"\n";
    }
}