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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#86662#5407. 基础图论练习题xzzduang0 0ms0kbC++142.0kb2023-03-10 14:54:042023-03-10 14:54:05

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-03-10 14:54:05]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2023-03-10 14:54:04]
  • 提交

answer

#include<iostream>
#include<stdio.h>
#include<ctype.h>
#include<string.h>
#define ll long long
#define ld long double
#define fi first
#define se second
#define pii pair<int,int>
#define lowbit(x) ((x)&-(x))
#define popcount(x) __builtin_popcount(x)
#define inf 0x3f3f3f3f
#define infll 0x3f3f3f3f3f3f3f3f
#define umap unordered_map
#define all(x) x.begin(),x.end()
#define mk make_pair
#define pb push_back
#define ckmax(x,y) x=max(x,y)
#define ckmin(x,y) x=min(x,y)
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
#define N 5005
using namespace std;
inline int read(){
	int x=0,f=0; char ch=getchar();
	while(!isdigit(ch)) f|=(ch==45),ch=getchar();
	while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return f?-x:x;
}
bool G[N][N];
int n,dfn[N],low[N],vis[N],st[N],top;
char s[N];
const int mo=1e9+7;
inline void red(int &x){x>=mo?x-=mo:0;}
inline int qpow(int x,int b){
	int res=1;
	for(;b;x=1LL*x*x%mo,b>>=1) if(b&1) res=1LL*res*x%mo;
	return res;
}
int res;
void tarjan(int u){
	dfn[u]=low[u]=++dfn[0];
	vis[u]=1,st[++top]=u;
	for(int v=1;v<=n;++v){
		if(!G[u][v]) continue;
		if(!dfn[v]){
			tarjan(v);
			ckmin(low[u],low[v]);
		}
		else if(vis[v]) ckmin(low[u],dfn[v]);
	}
	if(dfn[u]==low[u]){
		res++;
		while(st[top]!=u) vis[st[top]]=0,top--;
		vis[st[top]]=0;
		top--;
	}
}
inline int calc(){
	dfn[0]=0;
	for(int i=1;i<=n;++i) dfn[i]=low[i]=vis[i]=0;
	top=res=0;
	for(int i=1;i<=n;++i) if(!dfn[i]) tarjan(i);
	return res;
}
signed main(){
	n=read();
	for(int i=2;i<=n;++i){
		scanf("%s",s+1);
		int len=strlen(s+1);
		for(int j=1;j<=len;++j){
			int x;
			if(!isdigit(s[j])) x=s[j]-'A'+10;
			else x=s[j]-'0';
			for(int k=0;k<=3;++k){
				G[i][4*j+k-3]=((x>>k)&1);
			}
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=i+1;j<=n;++j) G[i][j]=(!G[j][i]);
	}
	int ans=0;
	for(int i=1;i<=n;++i){
		for(int j=1;j<i;++j){
			swap(G[i][j],G[j][i]);
			red(ans+=1LL*qpow(2,(i-2)*(i-1)/2+j-1)*calc()%mo);
			swap(G[i][j],G[j][i]);
		}
	}
	cout<<ans;
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Subtask #1:

score: 0
Runtime Error

Test #1:

score: 0
Runtime Error

input:

10000
100
1
2
2
8
C0
F0
27
78
AE1
C01
511
D87
EF20
3873
2742
73D0
DC9B0
FB2A3
9C011
9B4E0
95DC00
A7B980
F43531
6A6245
5347BE0
1A6C8A1
88E46D6
64CF3AE
D25F63C1
C894E4C3
1C0AFD73
EC1C3F9A
087CE17C0
22149A380
B28038AF1
B9CA21C7F
D78F5307C1
49045489A2
72C4DE6FD1
7713F40D05
EEE8878EEC1
310E62812B1
DA9D5B...

output:


result:


Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Skipped

Dependency #1:

0%

Subtask #4:

score: 0
Skipped

Dependency #1:

0%

Subtask #5:

score: 0
Skipped

Dependency #1:

0%