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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
#571439#9319. Bull Farmyukino_yukinoshitaTL 182ms11704kbC++143.4kb2024-09-17 23:05:132024-09-17 23:05:15

Judging History

你现在查看的是最新测评结果

  • [2024-09-17 23:05:15]
  • 评测
  • 测评结果:TL
  • 用时:182ms
  • 内存:11704kb
  • [2024-09-17 23:05:13]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

const int N = 2005;
const int M = 1e6 + 5;

int n, m, l, t[N][N];
int vis[N], flg[N], u1[N], u2[N], tt[N];
struct Qry {int a, b, c, id;} q[M];

bool cmp(Qry x, Qry y) {return x.c < y.c;}

bitset<2048> f[N];

int fa[N];
int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);}
void link(int x, int y)
{
  for(int i = 1; i <= n; i++)
    if(f[i][x]) f[i] |= f[y];
}
void merge(int x, int y)
{
  x = find(x), y = find(y);
  link(x, y), link(y, x);
  fa[y] = x;
}

bool vs[N];
bool g[N][N];

bool bfs(int s, int t)
{
  memset(vs, 0, sizeof(vs));
  queue<int> que;
  que.push(s); vs[s] = 1;
  while(!que.empty())
  {
    int u = que.front(); que.pop();
    // printf("u = %d\n", u);
    if(u == t) return 1;
    for(int i = 1; i <= n; i++) if(g[u][i] && !vs[i])
      que.push(i), vs[i] = 1;
  }
  return 0;
}

void solve()
{
  cin >> n >> l >> m;
  for(int i = 1; i <= l; i++)
  {
    string s; cin >> s;
    for(int j = 1; j <= n; j++) vis[j] = 0;
    int cnt = 0;
    for(int j = 1; j <= n; j++)
    {
      t[i][j] = 50 * (s[2 * j - 2] - '0') + s[2 * j - 1] - '0';
      // cout << t[i][j] << " \n"[j == n];
      if(vis[t[i][j]])
      {
        cnt++;
        if(cnt == 1) flg[i] = 1, u1[i] = j, u2[i] = vis[t[i][j]];
        else flg[i] = 2;
      }
      vis[t[i][j]] = j;
    }
    if(flg[i] == 1)
    {
      for(int j = 1; j <= n; j++) if(!vis[j])
      {
        tt[i] = j;
        break;
      }
    }
  }

  // cout << "\n";

  for(int i = 1; i <= m; i++)
  {
    string s; cin >> s;
    q[i].a = 50 * (s[0] - '0') + s[1] - '0';
    q[i].b = 50 * (s[2] - '0') + s[3] - '0';
    q[i].c = 50 * (s[4] - '0') + s[5] - '0';
    q[i].id = i;
    // cerr << q[i].a << " " << q[i].b << " " << q[i].c << "\n";
  }
  sort(q + 1, q + m + 1, cmp);
  memset(vis, 0, sizeof(vis));
  for(int i = 1; i <= n; i++) f[i][i] = 1, fa[i] = i;
  memset(g, 0, sizeof(g));

  string ans; ans.resize(m);
  for(int i = 1, cc = 1; i <= m; i++)
  {
    for(; cc <= q[i].c; cc++)
    {
      // printf("cc = %d q = %d\n", cc, q[i].c);
      if(flg[cc] == 2) continue;
      if(flg[cc] == 1)
      {
        // printf("edge1 %d %d\n", u1[cc], tt[cc]);
        // printf("edge2 %d %d\n", u2[cc], tt[cc]);
        g[u1[cc]][tt[cc]] = 1;
        g[u2[cc]][tt[cc]] = 1;
        continue;
      }
      for(int u = 1; u <= n; u++) if(vis[u] == 0)
      {
        int st = u;
        bool ok = 0;
        bitset<2048> to, from;
        for(; vis[u] == 0; u = t[cc][u])
        {
          vis[u] = 1, to |= f[u], from[u] = 1;
          if(find(u) != find(st)) merge(st, u);
          // printf("edge3 %d %d\n", st, u);
          // printf("edge3 %d %d\n", u, st);
          g[st][u] = g[u][st] = 1;
        }
      }
      memset (vis, 0, sizeof(vis));
    }
    // printf("cc = %d\n",)
    // ans[q[i].id - 1] = f[q[i].a][q[i].b] + '0';
    // printf("%d %d\n", q[i].a, q[i].b);
    // for(int I=1; I<=n; ++I) for(int J=1; J<=n; ++J) printf("%d%c", g[I][J], J==n?10:32);
    ans[q[i].id - 1] = bfs(q[i].a, q[i].b) + '0';
  }
  cout << ans << "\n";

  for(int i = 1; i <= n; i++) f[i].reset();
  for(int i = 1; i <= l; i++) flg[i] = 0;
  return;
}

int main()
{
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  int T; cin >> T;
  while(T--) solve();
  return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 11704kb

input:

2
5 2 4
0305040201
0404040404
030300
020500
050102
020501
6 2 4
030603010601
010203060504
030202
060402
050602
060401

output:

1011
0100

result:

ok 2 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 10524kb

input:

1
3 3 6
020202
030301
030201
020102
030203
010201
010303
020303
010202

output:

010101

result:

ok single line: '010101'

Test #3:

score: 0
Accepted
time: 182ms
memory: 10792kb

input:

200
10 10 5000
01060:04020305080709
0103070:060204050908
09070503080401060:02
050308010204090:0607
03010502040607080:09
03080109020504060:07
06050:09040302080107
07080305010409060:02
030809010:0204060507
0:060908070201050304
060700
090:03
09080:
070405
010703
0:0100
080601
030600
070206
0:0:09
08040...

output:

011110001101101111111111111111111101111111110111011110110110111011010111111111111111111101111111111110111111110111111111111101111111111110111111111111111111110001100111111111111111111111111011101111111111111111111111111111111111111111011011110100111110111111110111111100111111101110111111111101111110...

result:

ok 200 lines

Test #4:

score: -100
Time Limit Exceeded

input:

1
2000 1 1000000
M=:]A@8UAY7W2JJ4KEHIA[HSCQ1ENSC`JXR;F3PJ:_@41P9Z=9HR8P<<:DUXRR9^WOQFL?NZP6S@=J0^WE32=6;\U0?88]Q_RNPUMT6YU<4<S]H?:7OCQYOT4YAV1^764ENWSDBED>M7A:BI>KSIR48JQ9B=N\5T3N4A2aF0@>3TI81<G7;YE>W`NMP<:IT4CI3D0=GZC3I\CLQJQBA9BDIS9SAM55KaVA<Z@D=>:Y?CQHUQ5U3a6UVI8OKX9_FAF^7=5M85;<0;8YPAM<7Z7PP7A=N...

output:


result: