• BZOJ 1046 上升序列(LIS变形)


    要保证长度为L的序列下标字典序最小,当然要尽量选前面的数。

    如何判断前面的数是否满足条件?,只需要知道这个数开头的递增序列的最长长度是多少,如果不小于L,那么必然可以加入这个数。还需判断一下它是否大于前面的那个数就行了。

    LIS用nlogn。

    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi acos(-1.0)
    # define eps 1e-9
    # define MOD 1000000007
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int res=0, flag=0;
        char ch;
        if((ch=getchar())=='-') flag=1;
        else if(ch>='0'&&ch<='9') res=ch-'0';
        while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
        return flag?-res:res;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=10005;
    //Code begin...
    
    int a[N], dp[N], len[N], num=0;
    
    int bin_sea(int x)
    {
        int l=1, r=num+1, mid;
        while (l<r) {
            mid=(l+r)>>1;
            if (len[mid]<=x) r=mid;
            else l=mid+1;
        }
        if (r==num+1) ++num;
        len[r]=x;
        return r;
    }
    int main ()
    {
        int n, m, l, ma=0;
        scanf("%d",&n);
        FOR(i,1,n) scanf("%d",a+i);
        for (int i=n; i>=1; --i) dp[i]=bin_sea(a[i]), ma=max(ma,dp[i]);
        scanf("%d",&m);
        while (m--) {
            scanf("%d",&l);
            if (l>ma) {puts("Impossible"); continue;}
            int pre=0;
            FOR(i,1,n) {
                if (dp[i]>=l&&a[i]>pre) printf(l==1?"%d":"%d ",a[i]), --l, pre=a[i];
                if (!l) break;
            }
            putchar('
    ');
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6498258.html
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