Educational Codeforces Round 43

Educational Codeforces Round 43 

A. Minimum Binary Number

显然可以把所有\(1\)合并成一个

注意没有\(1\)的情况

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

void solve(){
    int n;
    string s;
    cin >> n >> s;
    int c0 = 0, c1 = 0;
    for(int x : s) if(x=='0') c0++; else c1++;
    if(!c1) cout << 0 << endl;
    else{
        cout << 1;
        while(c0--) cout << 0;
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

B. Lara Croft and the New Game

就摁模拟它怎么走的就好了，有点小麻烦

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
typedef long long int LL;
void solve(){
    LL n, m, k;
    cin >> n >> m >> k;
    if(k<n) cout << k + 1 << ' ' << 1 << endl;
    else{
        k -= n;
        int num = k / (m-2+m);
        int lft = k % (m-2+m);
        int x = n - 2 * num;
        if(!lft) cout << x << ' ' << 2 << endl;
        else{
            if(lft<=m-2) cout << x << ' ' << 2 + lft << endl;
            else{
                lft -= m - 1;
                x -= 1;
                cout << x << ' ' << m - lft << endl;
            }
        }
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

C. Nested Segments

把所有线段先按左端点从小到大排序，左端点相同的按右端点从大到小排，然后扫一遍，这样就不用管左端点的大小关系了，只要维护遍历过程中右端点最右的位置和相应线段的编号即可

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 3e5+7;

int n;
pair<pair<int,int>,int> seg[MAXN];
void solve(){
    ____();
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> seg[i].first.first >> seg[i].first.second;
        seg[i].second = i;
    }
    sort(seg+1,seg+1+n,[&](pair<pair<int,int>, int> a, pair<pair<int,int>,int > b){
        if(a.first.first==b.first.first) return a.first.second > b.first.second;
        else return a.first.first < b.first.first;
    });
    int rm = 0, id = 0;
    for(int i = 1; i <= n; i++){
        if(seg[i].first.second>rm){
            rm = seg[i].first.second;
            id = seg[i].second;
        }else{
            cout << seg[i].second << ' ' << id << endl;
            return;
        }
    }
    cout << -1 << ' ' << -1 << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

D. Degree Set

一开始有\(d_n+1\)个点，要构造出度数集合为\(d_1,d_2,\cdots,d_n\)的图

把编号\(1-d_1\)的点向其他所有点连边，那么\(1-d_1\)的点的度数为\(d_n-1\)，其他点的度数为\(d_1\)

保留\(1-d_1\)的点和\(d_{n-1}+2-d_n+1\)的点

剩下的\(d_1+1-d_{n-1}+1\)的点（一共\(d_{n-1}-d_1+1\)个点），需要构造一个度数集合为\(d_2-d_1,d_3-d_1,\cdots,d_{n-1}-d_1\)的图，可以发现是一个小规模的原问题，递归解决即可

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e3+7;

void solve(){
    ____();
    static int n, d[MAXN]; cin >> n;
    for(int i = 1; i <= n; i++) cin >> d[i];
    vector<pair<int,int> > edges;
    int l = 1, r = n, D = 0;
    for(int l = 1, r = n, D = 0; l <= r; D += d[l], l++, r--){
        for(int i = 1; i <= d[l]; i++) for(int j = i + 1; j <= d[r] + 1; j++) edges.push_back({i+D,j+D});
        for(int i = l + 1; i <= r - 1; i++) d[i] -= d[l];
    }
    printf("%d\n",edges.size());
    for(auto &e : edges) printf("%d %d\n",e.first,e.second);
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

E. Well played!

显然把生命翻倍给同一个随从是最优的

那么枚举给的这个随从即可

注意没有心火(攻击等于生命)的情况

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
typedef long long int LL;
const int MAXN = 2e5+7;

int n, a, b;
pair<LL,LL> cre[MAXN];
LL pre[MAXN];
void solve(){
    ____();
    cin >> n >> a >> b;
    b = min(n,b);
    for(int i = 1; i <= n; i++) cin >> cre[i].first >> cre[i].second;
    sort(cre+1,cre+1+n,[&](pair<LL,LL> &a, pair<LL,LL> &b){
        return a.first - a.second > b.first - b.second;
    });
    for(int i = 1; i <= n; i++) pre[i] = pre[i-1] + max(0ll,cre[i].first - cre[i].second);
    LL sum = 0, ret = 0;
    for(int i = 1; i <= n; i++) sum += cre[i].second;
    ret = sum + pre[b];
    sum += pre[b];
    for(int i = 1; i <= b; i++){
        LL ts = sum - max(0ll,cre[i].first-cre[i].second);
        ts -= cre[i].second;
        ret = max(ret,ts+(cre[i].first<<a));
    }
    if(b){
        for(int i = b + 1; i <= n; i++){
            LL ts = sum - max(0ll,cre[b].first-cre[b].second);
            ts -= cre[i].second;
            ret = max(ret,ts+(cre[i].first<<a));
        }
    }
    cout << ret << endl;
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}

F. Minimal k-covering

考虑网络流建图，所有边从左半图连右半图，容量为\(1\)，源点向左半图所有点连边，容量为\(deg[x]-k\)，右半图的点向汇点连边，容量为\(deg[y]-k\)，然后有流量的边边即为需要删掉的边

从大到小枚举\(k\)，每次从源点到左半图的点增加\(1\)容量的边，右半图的点向汇点增加\(1\)容量的边，增广一次即可

view code

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 4e3+7;
const int INF = 0x3f3f3f3f;
#define S 0
#define T MAXN - 1
set<int> edges;
struct EDGE{
    int to,cap,rev,id;
    EDGE(){};
    EDGE(int to, int cap, int rev, int id):to(to),cap(cap),rev(rev),id(id){};
};
vector<EDGE> G[MAXN];
int iter[MAXN],rk[MAXN];
void ADDEDGE(int u, int v, int cap, int id){
    G[u].push_back(EDGE(v,cap,(int)G[v].size(),id));
    G[v].push_back(EDGE(u,0,(int)G[u].size()-1,-id));
}
bool bfs(){
    memset(rk,0,sizeof(rk));
    memset(iter,0,sizeof(iter));
    rk[S] = 1;
    queue<int> que;
    que.push(S);
    while(!que.empty()){
        int u = que.front();
        que.pop();
        for(auto e : G[u]){
            if(!e.cap or rk[e.to]) continue;
            rk[e.to] = rk[u] + 1;
            que.push(e.to);
        }
    }
    return rk[T]!=0;
}
int dfs(int u, int flow){
    if(u==T) return flow;
    for(int &i = iter[u]; i < (int)G[u].size(); i++){
        auto &e = G[u][i];
        if(!e.cap or rk[e.to]!=rk[u]+1) continue;
        int d = dfs(e.to,min(e.cap,flow));
        if(d){
            e.cap -= d;
            G[e.to][e.rev].cap += d;
            if(e.id){
                if(e.id>0) edges.erase(e.id);
                else edges.insert(-e.id);
            }
            return d;
        }
    }
    return 0;
}
void Dinic(){
    while(bfs()){
        int d = dfs(S,INF);
        while(d) d = dfs(S,INF);
    }
}
int n1, n2, m, deg1[MAXN], deg2[MAXN];
void solve(){
    ____();
    cin >> n1 >> n2 >> m;
    for(int i = 1; i <= m; i++){
        int u, v; cin >> u >> v;
        ADDEDGE(u,v+n1,1,i);
        deg1[u]++; deg2[v]++;
    }
    for(int i = 1; i <= m; i++) edges.insert(i);
    int mindeg = min(*min_element(deg1+1,deg1+1+n1),*min_element(deg2+1,deg2+1+n2));
    vector<vector<int> > vec(mindeg,vector<int>());
    for(int i = 1; i <= n1; i++) ADDEDGE(S,i,deg1[i]-mindeg,0);
    for(int i = 1; i <= n2; i++) ADDEDGE(i+n1,T,deg2[i]-mindeg,0);
    for(int k = mindeg; k; k--){
        Dinic();
        for(int e : edges) vec[k-1].push_back(e);
        for(int i = 1; i <= n1; i++) ADDEDGE(S,i,1,0);
        for(int i = 1; i <= n2; i++) ADDEDGE(i+n1,T,1,0);
    }
    puts("0");
    for(int i = 0; i < mindeg; i++){
        printf("%d",vec[i].size());
        for(int x : vec[i]) printf(" %d",x);
        puts("");
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Local.in","r",stdin);
    freopen("ans.out","w",stdout);
    #endif
    solve();
    return 0;
}