POJ 2104【整体二分】

墨蓝 2022-08-04 12:50 132阅读 0赞

整体二分的想法就是将操作进行分块，将相互影响的操作放在一块。

这题要注意分治结束条件  
在没有离散化的情况下，就要以操作判断结束条件，当然这题操作数比较小，这样的话也更快。  
下面的姿势是重排下标，总感觉重排结构体会慢一些….实际上好像没什么影响

// whn6325689
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    #define root 1,1,n
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    
    const int MAXN=100010;
    int n,m;
    
    struct Data
    {
        int val,pos;
        bool operator < (const Data& b) const
        {
       return val<b.val;}
    }a[MAXN];
    struct Query
    {
        int l,r,k;
    }q[5050];
    
    
    int c[MAXN],ans[MAXN];
    int pos;
    
    void add(int i,int val)
    {
        for(;i<n+10;i+=lowbit(i))
            c[i]+=val;
    }
    
    int getsum(int i)
    {
        int sum=0;
        for(;i;i-=lowbit(i))
            sum+=c[i];
        return sum;
    }
    
    int id[5050],tmp1[5050],tmp2[5050];
    
    void solve(int h,int t,int l,int r)
    {
        if(l==r || h>t)//没有离散化
        {
            return;
        }
        int mid=MID(l,r);
        while(a[pos+1].val<=mid && pos<n)
        {
            add(a[pos+1].pos,1);
            pos++;
        }
        while(a[pos].val>mid)
        {
            add(a[pos].pos,-1);
            pos--;
        }
        int l1=0,l2=0;
        for(int i=h;i<=t;i++)
        {
            if(getsum(q[id[i]].r)-getsum(q[id[i]].l-1) >= q[id[i]].k)
            {
                ans[id[i]]=mid;
                tmp1[l1++]=id[i];
            }
            else
                tmp2[l2++]=id[i];
        }
        memcpy(id+h,tmp1,sizeof(int)*l1);
        memcpy(id+h+l1,tmp2,sizeof(int)*l2);
        solve(h,h+l1-1,l,mid);
        solve(h+l1,t,mid+1,r);
    }
    
    int main()
    {
    
        while(scanf("%d %d",&n,&m)!=EOF)
        {
            CLR(c,0);pos=0;
            int minn=INF,maxx=-1;
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i].val);
                a[i].pos=i;
                maxx=max(a[i].val,maxx);
                minn=min(a[i].val,minn);
            }
            sort(a+1,a+n+1);
            for(int i=1;i<=m;i++)
            {
                scanf("%d %d %d",&q[i].l,&q[i].r,&q[i].k);
                id[i]=i;
            }
            solve(1,m,minn,maxx+1);
            for(int i=1;i<=m;i++)
                printf("%d\n",ans[i]);
        }
        return 0;
    }

离散化之后的写法  
这样会相对慢一些，因为数的范围更大一些

//      whn6325689
    //      Mr.Phoebe
    //      http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    #define root 1,1,n
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    
    const int MAXN=100010;
    int n,m;
    
    struct Data
    {
        int val,pos;
        bool operator < (const Data& b) const
        {
       return val<b.val;}
    }a[MAXN];
    struct Query
    {
        int l,r,k;
        int id;
    }q[5050],q1[5050],q2[5050];
    
    
    int c[MAXN],ans[5050],b[MAXN];
    int pos;
    
    void add(int i,int val)
    {
        for(;i<n+10;i+=lowbit(i))
            c[i]+=val;
    }
    
    int getsum(int i)
    {
        int sum=0;
        for(;i;i-=lowbit(i))
            sum+=c[i];
        return sum;
    }
    
    void solve(int h,int t,int l,int r)
    {
        if(l==r)
        {
            for(int i=h;i<=t;i++)
                ans[q[i].id]=b[l];
            return;
        }
        int mid=MID(l,r);
        while(a[pos+1].val<=mid && pos<n)
        {
            add(a[pos+1].pos,1);
            pos++;
        }
        while(a[pos].val>mid)
        {
            add(a[pos].pos,-1);
            pos--;
        }
        int l1=0,l2=0;
        for(int i=h;i<=t;i++)
        {
            if(getsum(q[i].r)-getsum(q[i].l-1) >= q[i].k)
                q1[l1++]=q[i];
            else
                q2[l2++]=q[i];
        }
        memcpy(q+h,q1,sizeof(Query)*l1);
        memcpy(q+h+l1,q2,sizeof(Query)*l2);
        solve(h,h+l1-1,l,mid);
        solve(h+l1,t,mid+1,r);
    }
    
    int main()
    {
    
        while(scanf("%d %d",&n,&m)!=EOF)
        {
            CLR(c,0);pos=0;
            int cnt=0;
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i].val);
                a[i].pos=i;
                b[++cnt]=a[i].val;
            }
            sort(b+1,b+cnt+1);
            cnt=unique(b,b+cnt+1)-b-1;
            for(int i=1;i<=n;i++)
            {
                a[i].val=lower_bound(b+1,b+cnt+1,a[i].val)-b;
            }
            sort(a+1,a+n+1);
            for(int i=1;i<=m;i++)
            {
                scanf("%d %d %d",&q[i].l,&q[i].r,&q[i].k);
                q[i].id=i;
            }
            solve(1,m,1,cnt);
            for(int i=1;i<=m;i++)
                printf("%d\n",ans[i]);
        }
        return 0;
    }