手速练习

Data	Problem	name	times	solution
10.28	POJ 1753	Flip Game	20min	枚举
10.29	POJ 3368	Frequent values	1h	RMQ
10.30		树的最大独立集	20min	树dp

poj1753

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iomanip> 
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 Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;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,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
typedef long long ll;
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;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
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;
} 
char s[5];
int a[5][5];
int b[5][5];
int main()
{
// freopen("poj1753.in","r",stdin);
// freopen(".out","w",stdout);

    Rep(i,4) {
        scanf("%s",s);
        Rep(j,4) 
            a[i][j]= (s[j] == 'b')?1:0;
    }
    int ans=100;
    Rep(i,1<<16)
    {
        memcpy(b,a,sizeof(b));
        int l=0;
        Rep(j,16)
        {
            int x=j/4,y=j%4;
            if (i&(1<<j)) {
                ++l;
                b[x][y]^=1;
                if (x<3) b[x+1][y]^=1;
                if (y<3) b[x][y+1]^=1;
                if (x>0) b[x-1][y]^=1;
                if (y>0) b[x][y-1]^=1;

            }
        }

        int t=0;
        Rep(x,4) Rep(y,4) t+=b[x][y];
        if (!t||t==16) {
            ans=min(ans,l); 
        } 

    }   

    if (ans<=16) cout<<ans<<endl; 
    else cout<<"Impossible"<<endl;
    return 0;
}

poj 3368

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iomanip>
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 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,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define N (100000) 
#define MAXN (500000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
int a[MAXN];
int L[MAXN],R[MAXN]; 
int h[MAXN];
double A[MAXN];
double f[MAXN][32];
int main()
{
// freopen("poj3368.in","r",stdin);
// freopen(".out","w",stdout);

    while(~scanf("%d%d",&n,&m) && n) {
        MEM(h)
        For(i,n) scanf("%d",&a[i]),a[i]+=N+1,h[a[i]]++;
// For(i,n) A[i]=(double)0.1+h[a[i]]+(double)a[i]/100000.0;

        MEM(R) MEMI(L)


        int nn=2*N+1;
        For(i,nn) f[i][0]=(double)h[i];
        For(i,n) L[a[i]]=min(L[a[i]],i),R[a[i]]=max(R[a[i]],i); 

        for(int j=1;(1<<j)<=nn;j++) {
            for(int i=1;i+(1<<j)-1<=nn;i++) {
                f[i][j]=max(f[i][j-1],f[i+(1<<j)-(1<<j-1)][j-1]);               
            }
        }


        For(i,m) {
            int l,r;
            scanf("%d%d",&l,&r);
            int l2=a[l]+1,r2=a[r]-1;

            int ma2=min(r,R[a[r]])-max(l,L[a[r]])+1;
            int ma3=min(r,R[a[l]])-max(l,L[a[l]])+1;
            int ans=max(ma2,ma3);   

            if (l2<=r2) {
                int p=0,len=r2-l2+1;
                while ((1<<p+1)<=len) ++p;
                double ma = max(f[l2][p] , f[r2-(1<<p)+1][p] );
                ans=max(ans,(int)ma);
            }
            printf("%d\n",ans);
        }
    }



    return 0;
}

树的最大独立集

#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 Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;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,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
typedef long long ll;
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+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
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;
} 
#define MAXN (100000+10) 
vector<int> son[MAXN];
int n,d[MAXN]={0},f[MAXN],s[MAXN],gs[MAXN];
void dfs(int x,int fa){
    int m=son[x].size();
    Rep(i,m) {
        if (son[x][i]==fa) continue;
        f[son[x][i]]=x;
        dfs(son[x][i],x);
    }
    d[x]=max(s[x],1+gs[x]); 
    if (fa)
    {
        s[fa] += d[x];
        if (f[fa]) gs[f[fa]]+=d[x];
    }
}

int main()
{
    freopen("树的最大独立集.in","r",stdin);
// freopen("树的最大独立集.out","w",stdout);

    cin>>n;
    For(i,n-1){
        int x,y;
        x=read(),y=read();
        son[x].pb(y);
        son[y].pb(x);
    }
    dfs(1,0);
    cout<<d[1]<<endl;

    return 0;
}