2020牛客暑期多校训练营（第三场）
| A | Clam and Fish |
| B | Classical String Problem |
| C |
| 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\n", 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",n)
#define slf(n) scanf("lld",&n)
#define plf(n) printf("lld\n",&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",n)
#define slf(n) scanf("lld",&n)
#define plf(n) printf("lld\n",&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\n",ans);
}
}
| I | Sorting the Array |
| J | Operating on the Tree |
| K | Eleven Game |
| L | Problem L is the Only Lovely Problem |
