2020牛客暑期多校训练营（第三场）

A	Clam and Fish

B	Classical String Problem

C	Operation Love

D	Points Construction Problem

E	Two Matchings

我的我的锅，这道题没看，后期约等于挂机了

题意：给你一个序列a，你要找到两种完全不同的整个序列的两两匹配，使得所有两两匹配差的绝对值的和最小，输出这个和

思考：因为需要两个不同序列的的绝对值最小，只需要：

第一个序列价格相邻最小的进行匹配即可

第二个序列只需要在4和6之间找最小的匹配即可

卡了一下cin

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#define rep(i,a,b) for(int i = a; i <= b ; i ++ )
#define pre(i,a,b) for(int i = b ; i >= a ; i --)
#define ll long long
#define inf 0x3f3f3f3f3f3f3fll
#define ull unsigned long long
#define ios ios::sync_with_stdio(false),cin.tie(0)
using namespace std;
typedef pair<int,int> PII ;
const int N=2e6+10,mod=998244353,p=233;

int t,n;
int a[N],f[N];

int main()
{
    ios;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        rep(i,1,n) scanf("%lld",&a[i]);
        sort(a+1,a+1+n);

        rep(i,1,n) f[i]=inf;
        f[0]=0;

        for(int i = 4; i <= n; i += 2)
        {
            if(i >= 6)
                f[i] = min(f[i], f[i - 6] + a[i] - a[i - 5]);
            f[i] = min(f[i], f[i - 4] + a[i] - a[i - 3]);
        }
        printf("%lld
", 2 * f[n]);
    }
    return 0;
}

　　

F	Fraction Construction Problem

G	Operating on a Graph

题意：

 当时没有看这道题，当时估计是看了也不会，因为不熟悉并查集的操作，注意点：

1、将相同颜色的使用并查集连接起来就好
2、然后使用vector或者是list来记录相邻的点就好

/***
 *               ii.                                         ;9ABH,
 *              SA391,                                    .r9GG35&G
 *                                             i3X31i;:,rB1
 *              iMs,:,i5895,                         .5G91:,:;:s1:8A
 *               33::::,,;5G5,                     ,58Si,,:::,sHX;iH1
 *                Sr.,:;rs13BBX35hh11511h5Shhh5S3GAXS:.,,::,,1AG3i,GG
 *                .G51S511sr;;iiiishS8G89Shsrrsh59S;.,,,,,..5A85Si,h8
 *               :SB9s:,............................,,,.,,,SASh53h,1G.
 *            .r18S;..,,,,,,,,,,,,,,,,,,,,,,,,,,,,,....,,.1H315199,rX,
 *          ;S89s,..,,,,,,,,,,,,,,,,,,,,,,,....,,.......,,,;r1ShS8,;Xi
 *        i55s:.........,,,,,,,,,,,,,,,,.,,,......,.....,,....r9&5.:X1
 *       59;.....,.     .,,,,,,,,,,,...        .............,..:1;.:&s
 *      s8,..;53S5S3s.   .,,,,,,,.,..      i15S5h1:.........,,,..,,:99
 *      93.:39s:rSGB@A;  ..,,,,.....    .SG3hhh9G&BGi..,,,,,,,,,,,,.,83
 *      G5.G8  9#@@@@@X. .,,,,,,.....  iA9,.S&B###@@Mr...,,,,,,,,..,.;Xh
 *      Gs.X8 S@@@@@@@B:..,,,,,,,,,,. rA1 ,A@@@@@@@@@H:........,,,,,,.iX:
 *     ;9. ,8A#@@@@@@#5,.,,,,,,,,,... 9A. 8@@@@@@@@@@M;    ....,,,,,,,,S8
 *     X3    iS8XAHH8s.,,,,,,,,,,...,..58hH@@@@@@@@@Hs       ...,,,,,,,:Gs
 *    r8,        ,,,...,,,,,,,,,,.....  ,h8XABMMHX3r.          .,,,,,,,.rX:
 *   :9, .    .:,..,:;;;::,.,,,,,..          .,,.               ..,,,,,,.59
 *  .Si      ,:.i8HBMMMMMB&5,....                    .            .,,,,,.sMr
 *  SS       :: h@@@@@@@@@@#; .                     ...  .         ..,,,,iM5
 *  91  .    ;:.,1&@@@@@@MXs.                            .          .,,:,:&S
 *  hS ....  .:;,,,i3MMS1;..,..... .  .     ...                     ..,:,.99
 *  ,8; ..... .,:,..,8Ms:;,,,...                                     .,::.83
 *   s&: ....  .sS553B@@HX3s;,.    .,;13h.                            .:::&1
 *    SXr  .  ...;s3G99XA&X88Shss11155hi.                             ,;:h&,
 *     iH8:  . ..   ,;iiii;,::,,,,,.                                 .;irHA
 *      ,8X5;   .     .......                                       ,;iihS8Gi
 *         1831,                                                 .,;irrrrrs&@
 *           ;5A8r.                                            .:;iiiiirrss1H
 *             :X@H3s.......                                .,:;iii;iiiiirsrh
 *              r#h:;,...,,.. .,,:;;;;;:::,...              .:;;;;;;iiiirrss1
 *             ,M8 ..,....,.....,,::::::,,...         .     .,;;;iiiiiirss11h
 *             8B;.,,,,,,,.,.....          .           ..   .:;;;;iirrsss111h
 *            i@5,:::,,,,,,,,.... .                   . .:::;;;;;irrrss111111
 *            9Bi,:,,,,......                        ..r91;;;;;iirrsss1ss1111
 狗头真帅*/
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <list>
#define pub(n) push_back(n)
#define pob(n) pop_back(n)
#define sf(n) scanf("%d",&n)
#define pf(n) printf("%d
",n)
#define slf(n) scanf("lld",&n)
#define plf(n) printf("lld
",&n)
#define rep(i,a,b) for(int i = a; i <= b ; i ++ )
#define pre(i,a,b) for(int i = a ; i >= b ; i --)
#define ll long long
#define PII pair<int,int>
#define inf 0x3f3f3f3f3f3f3fll
#define ull unsigned long long
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
const int N=2e6+10,mod=1e9+7;
 
int f[N],t,i,n,m,x,y,q,u;
int find(int x)
{
    if (x!=f[x]) f[x]=find(f[x]);
     return f[x];
}
list<int> G[N];
int main()
{
    sf(t);
    while (t--)
    {
        sf(n),sf(m);
        for (i=0;i<n;i++)
        {
            f[i]=i;
            G[i].clear();
        }
        while (m--)
        {
            sf(x),sf(y);
            G[x].push_back(y);
            G[y].push_back(x);
        }
        sf(q);
        while (q--)
        {
            sf(u);
            if (u!=find(u)) continue;
            list<int> s;
            for (auto v:G[u])
            {
                x=find(v);
                if (x==u) continue;
                else f[x]=u;
                s.splice(s.end(),G[x]);
            }
            swap(s,G[u]);
        }
        for (i=0;i<n;i++) cout<<find(i)<<" ";
        cout<<endl;
    }
    return 0;
 
}

　　

H	Sort the Strings Revision

补这道题完全是因为我看到笛卡尔树的时候是懵的， 所以想学学，结果细节操作没有学会，只学会了使用板子（差点把我心态搞炸）

通过二分来解：每次找到最小的Pi，将数列分成两部分，再继续递归，知道只有一个为止，然后比较大小就好
然后对于快速求区间内pi的值，可以使用笛卡尔树

/***
 *               ii.                                         ;9ABH,
 *              SA391,                                    .r9GG35&G
 *                                             i3X31i;:,rB1
 *              iMs,:,i5895,                         .5G91:,:;:s1:8A
 *               33::::,,;5G5,                     ,58Si,,:::,sHX;iH1
 *                Sr.,:;rs13BBX35hh11511h5Shhh5S3GAXS:.,,::,,1AG3i,GG
 *                .G51S511sr;;iiiishS8G89Shsrrsh59S;.,,,,,..5A85Si,h8
 *               :SB9s:,............................,,,.,,,SASh53h,1G.
 *            .r18S;..,,,,,,,,,,,,,,,,,,,,,,,,,,,,,....,,.1H315199,rX,
 *          ;S89s,..,,,,,,,,,,,,,,,,,,,,,,,....,,.......,,,;r1ShS8,;Xi
 *        i55s:.........,,,,,,,,,,,,,,,,.,,,......,.....,,....r9&5.:X1
 *       59;.....,.     .,,,,,,,,,,,...        .............,..:1;.:&s
 *      s8,..;53S5S3s.   .,,,,,,,.,..      i15S5h1:.........,,,..,,:99
 *      93.:39s:rSGB@A;  ..,,,,.....    .SG3hhh9G&BGi..,,,,,,,,,,,,.,83
 *      G5.G8  9#@@@@@X. .,,,,,,.....  iA9,.S&B###@@Mr...,,,,,,,,..,.;Xh
 *      Gs.X8 S@@@@@@@B:..,,,,,,,,,,. rA1 ,A@@@@@@@@@H:........,,,,,,.iX:
 *     ;9. ,8A#@@@@@@#5,.,,,,,,,,,... 9A. 8@@@@@@@@@@M;    ....,,,,,,,,S8
 *     X3    iS8XAHH8s.,,,,,,,,,,...,..58hH@@@@@@@@@Hs       ...,,,,,,,:Gs
 *    r8,        ,,,...,,,,,,,,,,.....  ,h8XABMMHX3r.          .,,,,,,,.rX:
 *   :9, .    .:,..,:;;;::,.,,,,,..          .,,.               ..,,,,,,.59
 *  .Si      ,:.i8HBMMMMMB&5,....                    .            .,,,,,.sMr
 *  SS       :: h@@@@@@@@@@#; .                     ...  .         ..,,,,iM5
 *  91  .    ;:.,1&@@@@@@MXs.                            .          .,,:,:&S
 *  hS ....  .:;,,,i3MMS1;..,..... .  .     ...                     ..,:,.99
 *  ,8; ..... .,:,..,8Ms:;,,,...                                     .,::.83
 *   s&: ....  .sS553B@@HX3s;,.    .,;13h.                            .:::&1
 *    SXr  .  ...;s3G99XA&X88Shss11155hi.                             ,;:h&,
 *     iH8:  . ..   ,;iiii;,::,,,,,.                                 .;irHA
 *      ,8X5;   .     .......                                       ,;iihS8Gi
 *         1831,                                                 .,;irrrrrs&@
 *           ;5A8r.                                            .:;iiiiirrss1H
 *             :X@H3s.......                                .,:;iii;iiiiirsrh
 *              r#h:;,...,,.. .,,:;;;;;:::,...              .:;;;;;;iiiirrss1
 *             ,M8 ..,....,.....,,::::::,,...         .     .,;;;iiiiiirss11h
 *             8B;.,,,,,,,.,.....          .           ..   .:;;;;iirrsss111h
 *            i@5,:::,,,,,,,,.... .                   . .:::;;;;;irrrss111111
 *            9Bi,:,,,,......                        ..r91;;;;;iirrsss1ss1111
 Õ桪¡ªÊÖ¶¯¹·Í·*/
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <list>
#define pub(n) push_back(n)
#define pob(n) pop_back(n)
#define sf(n) scanf("%d",&n)
#define pf(n) printf("%d
",n)
#define slf(n) scanf("lld",&n)
#define plf(n) printf("lld
",&n)
#define rep(i,a,b) for(int i = a; i <= b ; i ++ )
#define pre(i,a,b) for(int i = a ; i >= b ; i --)
#define ll long long
#define PII pair<int,int>
#define inf 0x3f3f3f3f3f3f3fll
#define ull unsigned long long
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
// ull gethash(int i,int j) { return hs[j]-hs[i-1]*base[j-i+1]; }
// int Find(int x) {return fa[x]==x? x : fa[x]=Find(fa[x]);}
using namespace std;
const int MAXN=2e6+10;
const int MOD=1e9+7;
int st[MAXN],ls[MAXN],rs[MAXN];
ll p[MAXN],d[MAXN];
int r[MAXN];
ll sd,a,b,mod,ans,M;
int n;
void init(int n)
{
    st[0]=0;
    for(int i=0; i<n; i++)
    {
        int k=st[0];
        while(k>0&&p[st[k]]>p[i])
            k--;
        if(k)
            rs[st[k]]=i;
        if(k<st[0])
            ls[i]=st[k+1];
        st[++k]=i;
        st[0]=k;
    }
}//笛卡尔树
void dfs(int k,int left,int right,int rnk)
{
    if(p[k]==inf||left>=right)
    {
        for(int i=left; i<=right; i++)
            r[i]=rnk+(i-left);    //边界直接标记名次
        return;
    }
    dfs(ls[k],left,k,rnk+(p[k]%10>d[k])*(right-k));
    dfs(rs[k],k+1,right,rnk+(p[k]%10<d[k])*(k-left+1));//注意rnk的转移
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        scanf("%lld%lld%lld%lld",&sd,&a,&b,&mod);
        for(int i=0; i<n; i++)
            p[i]=i;
        for(int i=1; i<n; i++)
            swap(p[sd%(i+1)],p[i]),sd=(sd*a%mod+b)%mod;
        scanf("%lld%lld%lld%lld",&sd,&a,&b,&mod);
        for(int i=0; i<n; i++)
            d[i]=sd%10,sd=(sd*a%mod+b)%mod;
        for(int i=0; i<n; i++)
            if(p[i]%10==d[i])
                p[i]=inf;//如果无意义设为极大值
        init(n);//建树
        dfs(st[1],0,n,0);//dfs跑分治
        ans=0;
        M=1;
        for(int i=0; i<=n; i++)
            ans=(ans+((r[i])*M)%MOD)%MOD,M=(M*10000019)%MOD;//按题意哈希
        printf("%lld
",ans);
    }
}

　　

I	Sorting the Array

J	Operating on the Tree

K	Eleven Game

L	Problem L is the Only Lovely Problem