Atcoder Regular Contest 92 D Two Sequences F Two Faced Edges 两道神题

Atcoder 3943 Two Sequences
Atcoder 3945 Two Faced Edges

Atcoder 3943 Two Sequences

给两个长度为n的数组a,b,求(a[1]+b[1])^(a[1]+b[2])^...^(a[n]^b[n]).
不能直接暴力,我们逐位考虑.
如果说最终结果的第kkk个二进制位等于1" role="presentation" style="position: relative;">111,则我们首先把a[i],b[j]a[i],b[j]a[i],b[j]模2k+12k+12^{k+1},可以发现这样子并不影响答案.剩下的情况必然存在2k≤a[i]+b[j]<2×2k或3×2k≤a[i]+b[j]<4∗2k2k≤a[i]+b[j]<2×2k或3×2k≤a[i]+b[j]<4∗2k2^k\leq a[i]+b[j].
我们将操作后的b数组排序,对于每一个a[i]a[i]a[i]用二分或者尺取求出符合条件的b[j]b[j]b[j].
这样复杂度是28×log(n)×n28×log(n)×n28\times log(n) \times n.
可是还能优化.我们发现每一次对2k2k2^k取模的时候,只是前面一部分不取,后面一部分取,而且如果从大到小枚举kkk,可以达成每一次后面的数字最多到2k" role="presentation" style="position: relative;">2k2k2^k的两倍,因此bbb数组就是前面一半单调递增,后面一半也单调递增,我们找到中间点进行一次归并排序,复杂度为O(n)" role="presentation" style="position: relative;">O(n)O(n)O(n).如果再用尺取法优化,就可以使复杂度减小到O(n×28).O(n×28).O(n\times 28).

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){int x=0,f=1;char c=gc();for (;!isdigit(c);c=gc()) f^=c=='-';for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');return f?x:-x;}
template <typename mitsuha>
inline bool read(mitsuha &x){x=0;int f=1;char c=gc();for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';if (!~c) return 0;for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');return x=f?x:-x,1;}
template <typename mitsuha>
inline int write(mitsuha x){if (!x) return 0&pc(48);if (x<0) x=-x,pc('-');int bit[20],i,p=0;for (;x;x/=10) bit[++p]=x%10;for (i=p;i;--i) pc(bit[i]+48);return 0;}
inline char fuhao(){char c=gc();for (;isspace(c);c=gc());return c;}
}using namespace chtholly;
using namespace std;
const int yuzu=2e5;
typedef ll fuko[yuzu|10];
fuko a,b;int n;void merge_sort(ll *a,ll mod){
int pos=n+1,i;
for (i=1;i<=n;++i){if (pos>n&&a[i]>=mod) pos=i;a[i]&=mod-1;}inplace_merge(a+1,a+pos,a+n+1);
}int get(ll x){
ll ans=0;int i,kp=n;
for (i=1;i<=n;++i){for (;kp>0&&a[i]+b[kp]>=x;--kp);ans+=n-kp; }return ans&1;
}int main(){
int i,j;
for (int t=1,zxy=0;t--;){n=read();for (i=1;i<=n;++i) read(a[i]);for (i=1;i<=n;++i) read(b[i]);sort(a+1,a+n+1),sort(b+1,b+n+1);ll ans=0;for (i=61;~i;--i){merge_sort(a,2ll<<i),merge_sort(b,2ll<<i);ans|=(get(1ll<<i)^get(2ll<<i)^get(3ll<<i))<<i;}//printf("Case #%d: ",++zxy);write(ans),pl;}
}

此题原题hdu5270大家可以试着交一下.

Atcoder 3945 Two Faced Edges

给一张有向图,求每一条边反转之后会不会引起强连通分量个数的变化.
一看就知道显然这是一道瞎搞题.
那么对于每一条边所连的两个点u→vu→vu\to v来说,同时满足或者同时不满足以下两个条件时,不会引起强连通分量个数的变化.
1.u不经过这条边也能到达v. 2.v可以到达u.
对于第2个条件,我们O(n×m)O(n×m)O(n\times m)进行nnn次dfs" role="presentation" style="position: relative;">dfsdfsdfs即可.
而第一个条件怎么办?
我们先正过来dfsdfsdfs,对能够dfsdfsdfs到的点编上不同的号,然后再倒过来dfsdfsdfs,判断两次dfsdfsdfs编到的号是否是相等的,如果相等则uuu只能通过这条边到达v" role="presentation" style="position: relative;">vvv.
接下来把mm<script type="math/tex" id="MathJax-Element-2257">m</script>条边存下来都跑一遍就可以了.

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){int x=0,f=1;char c=gc();for (;!isdigit(c);c=gc()) f^=c=='-';for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');return f?x:-x;}
template <typename mitsuha>
inline bool read(mitsuha &x){x=0;int f=1;char c=gc();for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';if (!~c) return 0;for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');return x=f?x:-x,1;}
template <typename mitsuha>
inline int write(mitsuha x){if (!x) return 0&pc(48);if (x<0) x=-x,pc('-');int bit[20],i,p=0;for (;x;x/=10) bit[++p]=x%10;for (i=p;i;--i) pc(bit[i]+48);return 0;}
inline char fuhao(){char c=gc();for (;isspace(c);c=gc());return c;}
}using namespace chtholly;
using namespace std;
const int yuzu=2e5;
typedef int fuko[yuzu|10];
fuko u,v;int vis[2][yuzu>>6][yuzu>>6];
vector<int> lj[yuzu|10];void dfs(int id,int p,int u,int col){
if (vis[id][p][u]) return;
vis[id][p][u]=col;
for (int i:lj[u]) dfs(id,p,i,col);
}int main(){
int i,n=read(),m=read();
for (i=1;i<=m;++i){u[i]=read(),v[i]=read();lj[u[i]].push_back(v[i]);}
for (i=1;i<=n;++i){vis[0][i][i]=vis[1][i][i]=1;int now=0;for (int j:lj[i]) dfs(0,i,j,++now);if (lj[i].size()) for (int j=lj[i].size()-1;~j;--j) dfs(1,i,lj[i][j],j+1);}
for (i=1;i<=m;++i){puts((vis[0][u[i]][v[i]]!=vis[1][u[i]][v[i]])^(vis[0][v[i]][u[i]]>0)?"diff":"same");}
}

谢谢大家.