[2016 Changzhou No. 1 summer camp Day7], 2016 Changzhou day7

[2016 Changzhou No. 1 summer camp Day7], 2016 Changzhou day7

Sequence)
[Description]
Bucket has a sequence, which is initially empty. It inserts 1-n into the sequence sequentially, where I is inserted to the right of the current ai number (ai = 0 indicates inserting to the leftmost part of the sequence ). It wants you to help
It finds the final sequence.
[Input Data]
The first line is an integer n. N positive integers a1 ~ In the second row ~ An.
[Output Data]
Output n integers in a row to indicate the final sequence. Separate the numbers with a space.
[Example input]
5
0 1 1 0 3
[Sample output]
4 1 3 5 2
[Data Scope]
For 30% of data, n <= 1000.
For 70% of data, n <= 100000
For 100% of data, n <= ai <I.

 

Question

Insert from the back to the front. For a number, insert it to the current space. Use the line segment tree to maintain the subscript of the current space position.

 

#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int n;int a[1000005],ans[1000005];struct hh{    int l,r,val;    bool flag;};hh lt[5000005];void build(int root,int l,int r){    int mid;    lt[root].l=l;    lt[root].r=r;    if(l==r)    {        lt[root].val=1;        return;    }    mid=(l+r)>>1;    build(root<<1,l,mid);    build(root<<1|1,mid+1,r);    lt[root].val=lt[root<<1].val+lt[root<<1|1].val;}int query(int root,int now){    int mid;    if(lt[root].l==lt[root].r) return lt[root].l;    if(lt[root<<1].val>=now) return query(root<<1,now);    else return query(root<<1|1,now-lt[root<<1].val);}void del(int root,int pre){    if(lt[root].l<=pre&&lt[root].r>=pre) lt[root].val--;    if(lt[root].l==lt[root].r)    {        lt[root].flag=true;        return;    }    if(lt[root<<1].l<=pre&&lt[root<<1].r>=pre) del(root<<1,pre);    else if(lt[root<<1|1].l<=pre&&lt[root<<1|1].r>=pre) del(root<<1|1,pre);}int main(){    int i,j,pre;    freopen("sequence.in","r",stdin);    freopen("sequence.out","w",stdout);    scanf("%d",&n);    for(i=1;i<=n;i++) scanf("%d",&a[i]);    build(1,1,n);    for(i=n;i>=1;i--)    {        pre=query(1,a[i]+1);        ans[pre]=i;        del(1,pre);    }    for(i=1;i<=n;i++) printf("%d ",ans[i]);    fclose(stdin);    fclose(stdout);    return 0;}

 

 

 

 

 

Backpack)
[Description]
He has n items and a m-sized backpack. Each item has a small value. He hopes that you can help him find the maximum number of valuable items in his backpack.
[Input Data]
The first line has two integers, n and m. The next n rows have two integers, xi and wi, indicating the size and value of the I-th item.
[Output Data]
One integer in a row indicates the maximum value.
[Example input]
5 100
95 80
4 18
3 11
99 100
2 10
[Sample output]
101
[Data Scope]
For 20% of data, xi <= 1500.
For 30% of the data, wi <= 1500.
For 100% of the data, n <= 40, 0 <= m <= 10 ^ 18, 0 <= xi, wi <= 10 ^ 15.

Question

Half-fold search + Binary Search

 

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int n,tn,cnt,tot;long long m,ans=0;struct hh{    long long x,w;}a[45],q[2000005];void dfsa(int,long long,long long);void dfsb(int,long long,long long);long long search(long long);bool cmp(hh a,hh b){return a.x==b.x?a.w>b.w:a.x<b.x;}long long maxx(long long a,long long b){return a>b?a:b;}int main(){    int i,j;    freopen("pack.in","r",stdin);    freopen("pack.out","w",stdout);    scanf("%d%lld",&n,&m);    tn=n>>1;    for(i=1;i<=n;i++) scanf("%lld%lld",&a[i].x,&a[i].w);    dfsa(1,0,0);tot=1;    sort(q+1,q+cnt+1,cmp);    for(i=2;i<=cnt;i++)        if(q[i].x!=q[i-1].x) q[++tot]=q[i];    for(i=1;i<=tot;i++)        if(q[i].w<q[i-1].w) q[i].w=q[i-1].w;    dfsb(tn+1,0,0);    printf("%lld",ans);    fclose(stdin);    fclose(stdout);    return 0;}void dfsa(int pre,long long xx,long long ww){    if(pre>tn)    {        q[++cnt]=(hh){xx,ww};        return;    }    dfsa(pre+1,xx,ww);    if(xx+a[pre].x<=m) dfsa(pre+1,xx+a[pre].x,ww+a[pre].w);}void dfsb(int pre,long long xx,long long ww){    long long temp;    if(pre>n)    {        temp=search(m-xx);        ans=maxx(ans,ww+temp);        return;    }    dfsb(pre+1,xx,ww);    if(xx+a[pre].x<=m) dfsb(pre+1,xx+a[pre].x,ww+a[pre].w);}long long search(long long xx){    int l,r,mid;    if(xx<q[1].x) return 0;    l=1;r=tot;    while(l<r)    {        mid=l+r+1>>1;        if(q[mid].x<=xx) l=mid;        else r=mid-1;    }    return q[l].w;}

 

 

 

 

 

Segment tree)
[Description]
To redefine a line segment tree, the root node still represents [1, n], but the left and right sons of [l, r] represent [l, k] and [k + 1, r]. k can be used in [l, r.
Each access starts from the root node and ends the access only when the current node is a leaf node. Otherwise, the access will be directed down.
He will access m [ai, bi]. You need to help him select the appropriate k for each node and output the minimum sum of the number of nodes accessed each time.
[Input Data]
The first line has two integers, n and m, and the next line has two integers, ai and bi.
[Output Data]
An integer in a row indicates the answer.
[Example input]
6 6
1 4
2 6
3 4
3 5
2 3
5 6
[Sample output]
40
[Data Scope]
For 20% of data, n <= 50, m <= 100.
For 40% of data, n <= 200.
For 70% of data, n <= 1000.
For 100% of data, n <= 5000, m <= 100000.

Question

F [I, j] indicates the minimum number of accesses to the subtree of node [I, j]. w [I, j] indicates the node [I, j, j] the number of accesses, that is, the number of intervals that intersection [I, j.
F [I, j] = min {f [I, k] + f [k + 1, j]} + w [I, j] (I <= k <j)
First, w is pre-processed, so the complexity of this dp is O (n ^ 3), which can be optimized to O (n ^ 2) using a quadrilateral inequality)

 

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int n,m;int l[5005],r[5005],f[5005][5005],w[5005][5005];int main(){    int i,j,k,x,y;    freopen("segment.in","r",stdin);    freopen("segment.out","w",stdout);    scanf("%d%d",&n,&m);    for(i=1;i<=m;++i)    {        scanf("%d%d",&x,&y);        l[x]++;r[y]++;    }    for(i=1;i<=n;++i) r[i]+=r[i-1];    for(i=n;i>=1;i--) l[i]+=l[i+1];    memset(f,9999999,sizeof f);    for(i=1;i<=n;i++) f[i][i]=m-r[i-1]-l[i+1];    for(i=1;i<=n;i++) w[i][i]=i;    for(k=1;k<=n-1;k++)        for(i=1;i<=n-k;i++)        {            for(j=w[i][i+k-1];j<=w[i+1][i+k]&&j+1<=i+k;++j)                if(f[i][j]+f[j+1][i+k]<f[i][i+k])                {                    f[i][i+k]=f[i][j]+f[j+1][i+k];                    w[i][i+k]=j;                }            f[i][i+k]+=(m-r[i-1]-l[i+k+1]);        }    printf("%d",f[1][n]);    fclose(stdin);    fclose(stdout);    return 0;}