Codeforces Round #419 (Div. 1) (ABCD)

1. 815A Karen and Game

大意: 给定$nm$矩阵, 每次选择一行或一列全部减$1$, 求最少次数使得矩阵全$0$

贪心, $n>m$时每次取一列, 否则取一行

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = 1e6+;

int n,m,a[][];

int op[N], p[N], cnt;

void add(int tp, int x) {

    ++cnt;

    op[cnt] = tp, p[cnt] = x;

    if (tp==) REP(i,,n) --a[i][x];

    else REP(i,,m) --a[x][i];

}

int main() {

    scanf("%d%d",&n,&m);

    int sum = ;

    REP(i,,n) REP(j,,m) scanf("%d",a[i]+j),sum+=a[i][j];

    while (sum) {

        if (n>m) {

            int ok = ;

            REP(i,,m) {

                int s = 1e9;

                REP(j,,n) s=min(s,a[j][i]);

                if (s) add(,i),sum-=n,ok=;

            }

            if (!ok) {

                REP(i,,n) {

                    int s = 1e9;

                    REP(j,,m) s=min(s,a[i][j]);

                    if (s) add(,i),sum-=m,ok=;

                }

                if (!ok) return puts("-1"),;

            }

        }

        else {

            int ok = ;

            REP(i,,n) {

                int s = 1e9;

                REP(j,,m) s=min(s,a[i][j]);

                if (s) add(,i),ok=,sum-=m;

            }

            if (!ok) {

                REP(i,,m) {

                    int s = 1e9;

                    REP(j,,n) s=min(s,a[j][i]);

                    if (s) add(,i),ok=,sum-=n;

                }

                if (!ok) return puts("-1"),;

            }

        }

    }

    printf("%d\n",cnt);

    REP(i,,cnt) printf("%s %d\n",op[i]==?"col":"row",p[i]);

}

2. 815B Karen and Test

大意: 有一个三角形的数表, 给定第一行, 下面每一行是上面相邻两数和或差, 求最底层的数是多少.

打表算一下每个数的贡献, 找下规律即可

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = ;

vector<int> f[N][N];

int n, a[N];

vector<int> add(vector<int> a, vector<int> b) {

    REP(i,,n) a[i]+=b[i];

    return a;

}

vector<int> sub(vector<int> a, vector<int> b) {

    REP(i,,n) a[i]-=b[i];

    return a;

}

int main() {

    scanf("%d",&n);

    REP(i,,n) REP(j,,n) f[i][j].resize(n+);

    REP(i,,n) f[][i][i] = ;

    int cur = ;

    REP(i,,n) {

        REP(j,,n-i+) {

            if (cur) f[i][j]=add(f[i-][j],f[i-][j+]);

            else f[i][j]=sub(f[i-][j],f[i-][j+]);

            cur ^= ;

        }

    }

    REP(i,,n) printf("%d,",f[n][][i]);hr;

}

打表的代码

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = 1e6+;

int n, a[N], fac[N], ifac[N];

int f[N], g[N];

ll C(int n, int m) {

    if (n<m) throw;

    return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P;

}

int main() {

    fac[]=;

    REP(i,,N-) fac[i]=(ll)fac[i-]*i%P;

    ifac[N-]=inv(fac[N-]);

    PER(i,,N-) ifac[i]=(ll)ifac[i+]*(i+)%P;

    scanf("%d", &n);

    REP(i,,n) scanf("%d",a+i);

    if (n==) return printf("%d\n",a[]),;

    if (n%==) {

        int x = n/-;

        REP(i,,n) g[i]=i&?C(x,i/):-C(x,i/-);

    }

    else if (n%==) {

        int x = (n-)/;

        REP(i,,n) if (i&) g[i] = C(x,i/);

    }

    else if (n%==) {

        int x = (n-)/;

        REP(i,,n) g[i]=C(x,(i-)/);

    }

    else {

        int x = (n-)/;

        REP(i,,n-) f[i]=C(x,(i-)/);

        REP(i,,n) g[i]=i&?f[i]-f[i-]:f[i]+f[i-];

    }

    int ans = ;

    REP(i,,n) ans = (ans+(ll)g[i]*a[i])%P;

    if (ans<) ans += P;

    printf("%d\n", ans);

}

3. 815C Karen and Supermarket

大意: $n$个商品构成一棵树, 商品原价$c_i$元, 用优惠券会减少$d_i$元, 但如果$i$使用优惠券, 那么必须购买$x_i$. 预算为$b$元, 求最多买多少商品.

暴力树形$dp$即可, 复杂度是$O(n^2)$, 因为每个二元组只会在lca出被枚举到一次

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = 5e3+;

int n, b;

int c[N],d[N],fa[N],sz[N];

int f[N][N], g[N][N];

vector<int> a[N];

//f[x][y] = x子树内, x用优惠券, 一共购买y个的最少花费

//g[x][y] = x子树内, x不用优惠券, 一共购买y个的最少花费

void chkmin(int &a, int b) {a>b?a=b:;}

void dfs(int x) {

    f[x][] = g[x][] = ;

    for (int y:a[x]) {

        dfs(y);

        PER(i,,sz[x]) REP(j,,sz[y]) {

            chkmin(f[x][i+j],f[x][i]+min(g[y][j],f[y][j]));

            chkmin(g[x][i+j],g[x][i]+g[y][j]);

        }

        sz[x] += sz[y];

    }

    ++sz[x];

    PER(i,,sz[x]) {

        g[x][i] = min(g[x][i],g[x][i-]+c[x]);

        f[x][i] = f[x][i-]+c[x]-d[x];

    }

}

int main() {

    memset(f,0x3f,sizeof f);

    memset(g,0x3f,sizeof g);

    scanf("%d%d%d%d",&n,&b,c+,d+);

    REP(i,,n) {

        scanf("%d%d%d",c+i,d+i,fa+i);

        a[fa[i]].pb(i);

    }

    dfs();

    PER(i,,n) if (f[][i]<=b||g[][i]<=b) return printf("%d\n",i),;

}

4. 815D Karen and Cards

大意: $n$张卡, 每张卡三个属性, 上限分别为$p,q,r$, 若一张卡存在两个属性值严格大于另一张卡, 那么这张卡能打败另一张卡. 求有多少张卡能打败其他所有卡.

按$c$排序, 从大到小枚举$c$, 对$a,b$建一个二维平面, 那么假设一张卡的$c_i<c$, 那么这张卡贡献就是对$1\le x\le a_i,1\le y\le b_i$的矩形赋零. 若$c_i\ge c$, 那么相当于对$1\le x\le p,1\le y\le b_i$和$1\le x\le a_i,1\le y\le q$的两个矩形赋零. 用线段树区间取max, 查询最大值即可.

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+;

#else

const int N = 1e2+;

#endif

int n, p, q, r;

struct _ {int a,b,c;} f[N];

//区间取max, 区间求和

//维护区间min, 转化为区间赋值, 区间求和

struct {

    int ma,mi,tag;

    ll sum;

    void upd(int x, int t) {

        ma=mi=tag=x, sum=(ll)t*x;

    }

} tr[N<<];

void pu(int o) {

    tr[o].ma = max(tr[lc].ma,tr[rc].ma);

    tr[o].mi = min(tr[lc].mi,tr[rc].mi);

    tr[o].sum = tr[lc].sum+tr[rc].sum;

    tr[o].tag = ;

}

void pd(int o, int l, int r) {

    if (tr[o].tag) {

        tr[lc].upd(tr[o].tag,mid-l+);

        tr[rc].upd(tr[o].tag,r-mid);

        tr[o].tag = ;

    }

}

void upd(int o, int l, int r, int ql, int qr, int v) {

    if (ql>qr||tr[o].mi>=v) return;

    if (ql<=l&&r<=qr&&tr[o].ma<=v) return tr[o].upd(v,r-l+);

    pd(o,l,r);

    if (mid>=ql) upd(ls,ql,qr,v);

    if (mid<qr) upd(rs,ql,qr,v);

    pu(o);

}

int main() {

    scanf("%d%d%d%d", &n, &p, &q, &r);

    REP(i,,n) scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].c);

    REP(i,,n) upd(,,p,,f[i].a,f[i].b);

    sort(f+,f++n,[](_ a,_ b){return a.c>b.c;});

    ll ans = ;

    int now = ;

    PER(i,,r) {

        while (now<=n&&f[now].c>=i) {

            upd(,,p,,f[now].a,q);

            upd(,,p,f[now].a+,p,f[now].b);

            ++now;

        }

        ans += (ll)p*q-tr[].sum;

    }

    printf("%lld\n", ans);

}

5. 815E Karen and Neighborhood

大意: $n$个房间,$k$个人轮流去住, 第一个人住$1$号, 之后每个人会住距离有人的房间最远的房间, 有多个的话住编号最小的, 求第$k$个人房间号.

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