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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#464278#5523. Graph Problem With Small $n$BalintRTL 0ms3788kbC++202.2kb2024-07-06 00:02:422024-07-06 00:02:42

Judging History

你现在查看的是最新测评结果

  • [2024-07-06 00:02:42]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3788kb
  • [2024-07-06 00:02:42]
  • 提交

answer

#include <bits/stdc++.h>
#include <immintrin.h>
using namespace std;

typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpii;
typedef complex<double> cpx;
template <typename T> using minPq = priority_queue<T, vector<T>, greater<T>>;
#define ms(a, x) memset(a, x, sizeof(a))
#define pb push_back
#define fs first
#define sn second
#define ALL(v) begin(v), end(v)
#define SZ(v) ((int) (v).size())
#define lbv(v, x) (lower_bound(ALL(v), x) - (v).begin())
#define ubv(v, x) (upper_bound(ALL(v), x) - (v).begin())
template <typename T> inline void UNIQUE(vector<T> &v){sort(ALL(v)); v.resize(unique(ALL(v)) - v.begin());}
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
#define FR(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORR(i, a, b) for(int i = (a); i >= (b); i--)
#define dbg(x) {cerr << #x << ' ' << x << endl;}
#define dbgArr(arr, n) {cerr << #arr; FR(_i, n) cerr << ' ' << (arr)[_i]; cerr << endl;}
template <typename T, typename U>
ostream& operator<<(ostream &os, pair<T, U> p){return os << "(" << p.fs << ", " << p.sn << ")";}

const int MN = 24;
int n;
int adjMat[MN], iAdjMat[MN], ans[MN];
int dp[1<<(MN-1)];

#define proc(n1) dp[s1 | (1 << n1)] |= adjMat[n1] & -((inc >> n1) & 1);

void run(){
    dp[0] = adjMat[n-1];
    FR(s1, 1<<(n-1)){
        int inc = dp[s1] &= ~s1;
        FR(n1, n-1) proc(n1);
    }
}

void solve(){
    FR(src, n){
        vi ord(n);
        iota(ALL(ord), 0);
        swap(ord[src], ord[n-1]);
        ms(adjMat, 0);
        FR(i, n) FR(j, n) adjMat[i] |= ((iAdjMat[ord[i]] >> ord[j]) & 1) << j;
        fill_n(dp, 1<<(n-1), 0);
        run();
        int mask = (1<<(n-1)) - 1;
        FR(n1, n) if(n1 != src) ans[src] |= ((dp[mask ^ (1 << ord[n1])] >> ord[n1]) & 1) << n1;
    }
}

int main(){
    cin.sync_with_stdio(0); cin.tie(0);
    cin >> n;
    FR(i, n){
        string str; cin >> str;
        FR(j, n) iAdjMat[i] |= (str[j]-'0') << j;
    }

    solve();

    FR(i, n){
        FR(j, n) cout << ((ans[i] >> j) & 1);
        cout << '\n';
    }
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3616kb

input:

4
0110
1010
1101
0010

output:

0001
0001
0000
1100

result:

ok 4 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 3652kb

input:

6
010001
101000
010100
001010
000101
100010

output:

010001
101000
010100
001010
000101
100010

result:

ok 6 lines

Test #3:

score: 0
Accepted
time: 0ms
memory: 3788kb

input:

4
0111
1011
1101
1110

output:

0111
1011
1101
1110

result:

ok 4 lines

Test #4:

score: -100
Time Limit Exceeded

input:

23
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
00000000000000000000000
000000000...

output:


result: