Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined)

于 2017-10-07 10:12:29 发布
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分类专栏： 比赛题解
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本文链接：https://blog.youkuaiyun.com/nike0good/article/details/78168330
本文深入探讨了多种竞赛编程技巧，包括字符串匹配、时间计算及贪心算法的应用，并介绍了一种结合莫队算法与四边形不等式的高效解决方案。

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A. Bark to Unlock

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int main()
{
//  freopen("a.in","r",stdin);
//  freopen(".out","w",stdout);
    char s[100];
    scanf("%s",s);
    int n=read();
    bool fl=0,fl1=0,fl2=0;
    For(i,n) {
        char s2[100];
        scanf("%s",s2);
        if (strcmp(s,s2)==0) fl=1;
        if (s[1]==s2[0]) fl1=1;
        if (s[0]==s2[1]) fl2=1; 
    }
    if (fl || (fl1&&fl2)) {
        puts("YES");
    }else puts("NO");

    return 0;
}

B. Race Against Time

石英钟坐标可能不在整点

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);
    double h=5*read()%60,m=read(),s=read();
    h+=m/60/5+s/60/60/5;
    m+=s/60;

    int t1=read(),t2=read();
    t1=t1*5%60;
    t2=t2*5%60;
    if(t1>t2) swap(t1,t2);
    bool fl=0,fl2=0;
    if (t1<h&&h<t2) fl=1;
    if (t1<m&&m<t2) fl=1;
    if (t1<s&&s<t2) fl=1;

    if (!(t1<=h&&h<=t2)) fl2=1;
    if (!(t1<=s&&s<=t2)) fl2=1;
    if (!(t1<=m&&m<=t2)) fl2=1;
    puts((fl&&fl2) ?"NO":"YES");

    return 0;
}

C. Qualification Rounds

贪心发现最多只要取2场比赛

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int c[100]={};
int main()
{
//  freopen("c.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read(),k=read();
    Rep(i,n) {
        int j=0;
        Rep(l,k) j=j*2+(1^read());
        c[j]++;
    }
    int S=1<<k;S--;
    if (c[S]) puts("YES");
    else {
        Rep(i,S) Rep(j,S) if (i!=j &&c[i]&&c[j] &&((i|j)==S)) {
            puts("YES");return 0;
        }
        puts("NO");
    }

    return 0;
}

F. Yet Another Minimization Problem

莫队+四边形不等式
通过四边形不等式证明cost(i,j)满足四边形不等式
k题n很小，把模型变为，f(i)=f(j)+cost(i,j)
显然可以利用单调性 + 二分转移 。
计算的时候使用莫队，在本题中莫队保证线性。

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<algorithm>  
#include<functional>  
#include<iostream>  
#include<cmath>  
#include<cstring>  
#include<cctype>  
#include<ctime>  
using namespace std;  
#define For(i,n) for(int i=1;i<=n;i++)  
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)  
#define Forp(x) for(int p=pre[x];p;p=next[p])  
#define MAXN (100020+10)  
typedef long long ll;
int a[MAXN];  
int b[MAXN]={0};
int h[MAXN]={},tcase=0;  
long long Ti2=0;  
int n,m;
int ll1=1,rr1=0;
ll calc(int ll2,int rr2) {
    if (ll1<ll2) Fork(j,ll1,ll2-1) b[a[j]]*=(h[a[j]]==tcase),h[a[j]]=tcase,Ti2-=(long long)b[a[j]]*(b[a[j]]-1)/2,b[a[j]]--,Ti2+=(long long)b[a[j]]*(b[a[j]]-1)/2;  
    if (ll2<ll1) Fork(j,ll2,ll1-1) b[a[j]]*=(h[a[j]]==tcase),h[a[j]]=tcase,Ti2-=(long long)b[a[j]]*(b[a[j]]-1)/2,b[a[j]]++,Ti2+=(long long)b[a[j]]*(b[a[j]]-1)/2;  
    if (rr1<rr2) Fork(j,rr1+1,rr2) b[a[j]]*=(h[a[j]]==tcase),h[a[j]]=tcase,Ti2-=(long long)b[a[j]]*(b[a[j]]-1)/2,b[a[j]]++,Ti2+=(long long)b[a[j]]*(b[a[j]]-1)/2;  
    if (rr2<rr1) Fork(j,rr2+1,rr1) b[a[j]]*=(h[a[j]]==tcase),h[a[j]]=tcase,Ti2-=(long long)b[a[j]]*(b[a[j]]-1)/2,b[a[j]]--,Ti2+=(long long)b[a[j]]*(b[a[j]]-1)/2;  
    ll1=ll2,rr1=rr2;
    return Ti2;
}
void reset() {
    ll1=1,rr1=0;
    ++tcase;Ti2=0;
}
#define MAXK (30)
ll f[MAXN][MAXK];
void dfs(int k,int l,int r,int ll1,int rr1) {
    int i=(l+r)/2;
    f[i][k]=1e16;
    int LL=max(ll1,k-1),RR=min(rr1,i-1);
    int t=-1;
    ForkD(j,LL,RR) {
        if (f[i][k]> f[j][k-1] + calc(j+1,i)) t=j;
        f[i][k] = min(f[i][k], f[j][k-1] + calc(j+1,i));
    }
    int m=i;
    if (l<m) dfs(k,l,m-1,ll1,t);
    if (m+1<=r) dfs(k,m+1,r,t,rr1);
}
int main()  
{  
//  freopen("d.out","r",stdin);  
//  freopen("f.out","w",stdout);  
    scanf("%d%d",&n,&m);  
    For(i,n) scanf("%d",&a[i]);
    For(i,n) For(j,m) f[i][j]=1e15;

    For(i,n) f[i][1]=calc(1,i);

    Fork(k,2,m) {
        bool fl=0;
        dfs(k,k,n,1,n);
    }
    cout<<f[n][m]<<endl;
    return 0;  
}
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