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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
#577450#9319. Bull FarmfanhuaxingyuWA 57ms5876kbC++206.2kb2024-09-20 11:31:552024-09-20 11:31:57

Judging History

你现在查看的是最新测评结果

  • [2024-09-20 11:31:57]
  • 评测
  • 测评结果:WA
  • 用时:57ms
  • 内存:5876kb
  • [2024-09-20 11:31:55]
  • 提交

answer

#include <bits/stdc++.h>
#include<array>
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//#define int long long
#define lowbit(x) ((x) & (-x))
#define ar(x) array<int, x>
#define endl '\n'
#define all(x)  begin(x),end(x)
#define all2(x)  begin(x)+1,end(x)
using namespace std;
using i64 = long long;
const int N = 2e5 + 10;
const int M = 5e5 + 1;
const int INF = 0x3f3f3f3f;
const ll INF2 = 0x3f3f3f3f3f3f3f3f;
const long double PI = 3.1415926535897932384626;
const long double eps = 1e-5;
const ll mod = 19260817;
const ll mod2 = 998244353;
const int base = 131;
const int base2 = 13331;
int dx[] = { 1,-1,0,0 }, dy[] = { 0,0,1,-1 };
int dx2[] = { 2,-2,0,0 }, dy2[] = { 0,0,2,-2 };
ll qpow(ll a, ll b, ll p)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans *= a, ans %= p;
        a *= a;
        a %= p;
        b >>= 1;
    }
    return ans % p;
}
ll gcd(ll a, ll b)
{
    return !b ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}
int n, m;
//tim为时间
int dfn[N], low[N], tim = 0, vis[N], cnt = 0, top = 0;//dfn标记dfs遍历的顺序,low标记当前点通过某条边所能回溯到的最早的那个点,vis标记当前点是否在栈内
vector<int>g[2020], sg[2020];//邻接表存图和缩点后的图
int s[N];//栈
int scc[N];//缩点后原点属于缩点后的哪个点
void dfs(int x)
{
    s[++top] = x;
    vis[x] = true;//标记为在栈内
    dfn[x] = low[x] = ++tim;//时间入和出的初标记
    for (int i = 0; i < g[x].size(); ++i)
    {
        int y = g[x][i];
        if (dfn[y] == 0)//如果当前点没被遍历
        {
            dfs(y);//遍历当前当前点
            low[x] = min(low[x], low[y]);//更新low
        }
        else if (vis[y] == true)//如果当前点在栈内则更新low
        {
            low[x] = min(low[x], low[y]);
        }
        //如果不在栈内则不用管
    }
    if (dfn[x] == low[x])//如果当前二者相等
    {
        int t;
        ++cnt;
        do {
            t = s[top--];
            vis[t] = 0;
            scc[t] = cnt;
        } while (t != x);
    }
}
void tarjan()
{
    for (int i = 1; i <= n; ++i)//循环dfs,因为可能不连通
    {
        if (dfn[i] == 0)
            dfs(i);
    }
}
void init()
{
    top = tim = cnt = 0;
    for (int i = 1; i <= n; ++i)
    {
        g[i].clear();
        sg[i].clear();
        dfn[i] = scc[i] = low[i] = vis[i] = 0;
    }
}
void solve()
{
    int l, q;
    cin >> n >> l >> q;
    init();
    vector<vector<int>>dis(n + 1, vector<int>(n + 1, INF));
    for (int i = 1; i <= n; ++i)
        dis[i][i] = 0;
    for (int i = 1; i <= l; ++i)
    {
        string s;
        cin >> s;
        vector<int>cn(n + 1);
        vector<int>t(n + 1);
        for (int j = 1; j <= n; ++j)
        {
            t[j] = (s[2 * j - 2] - 48) * 50 + (s[2 * j - 1] - 48);
            cn[t[j]]++;
        }

        int ok = 0;
        for (int j = 1; j <= n; ++j)
        {
            if (cn[j] == 1)
            {
                ok++;
            }
        }
        if (ok == n)
        {
            for (int j = 1; j <= n; ++j)
            {
                if (dis[j][t[j]] == INF)
                    g[j].push_back(t[j]);
                dis[j][t[j]] = min(dis[j][t[j]], i);
                if (dis[t[j]][j] == INF)
                    g[t[j]].push_back(j);
                dis[t[j]][j] = min(dis[t[j]][j], i);
            }

        }
        else if (ok == n - 2)
        {
            vector<int>tmp;
            int tmp2 = 0;
            for (int j = 1; j <= n; ++j)
                if (cn[t[j]] == 2)
                    tmp.push_back(j);
            for (int j = 1; j <= n; ++j)
                if (cn[j] == 0)tmp2 = j;
            if (dis[tmp[0]][tmp2] == INF)
                g[tmp[0]].push_back(tmp2);
            if (dis[tmp[1]][tmp2] == INF)
                g[tmp[1]].push_back(tmp2);
            dis[tmp[0]][tmp2] = min(dis[tmp[0]][tmp2], i);
            dis[tmp[1]][tmp2] = min(dis[tmp[1]][tmp2], i);
        }
    }
    tarjan();
    vector<array<int, 3>>e;
    for (int i = 1; i <= n; ++i)
    {
        for (int v : g[i])
        {
            e.push_back({ dis[i][v],i,v });
        }
    }
    sort(all(e));
    vector<int>fa(n + 1);
    for (int i = 1; i <= n; ++i)
    {
        fa[i] = i;
    }
    function<int(int)>find = [&](int x)
    {
        return fa[x] == x ? x : fa[x] = find(fa[x]);
    };
    vector<vector<int>>vis2(n + 1, vector<int>(n + 1));
    for (auto [w, u, v] : e)
    {
        if (scc[u] == scc[v])
        {
            int a = find(u);
            int b = find(v);
            if (a != b)
            {
                fa[a] = b;
                sg[a].push_back(b);
                sg[b].push_back(a);
            }
        }
        else
        {
            if (!vis2[scc[u]][scc[v]])
            {
                vis2[scc[u]][scc[v]] = 1;
                sg[u].push_back(v);
            }
        }
    }
    for (int i = 1; i <= n; ++i)
    {
        vector<int>vis3(n + 1);
        vis3[i] = 1;
        queue<int>q;
        q.push(i);
        while (q.size())
        {
            int u = q.front();
            q.pop();
            for (int v : sg[u])
            {
                if (!vis3[v])
                {
                    vis3[v] = 1;
                    dis[i][v] = min(dis[i][v], max(dis[i][u], dis[u][v]));
                    q.push(v);
                }
            }
        }
    }
    vector<int>ans;
    while (q--)
    {
        string s;
        cin >> s;
        int a, b, c;
        a = (s[0] - 48) * 50 + (s[1] - 48);
        b = (s[2] - 48) * 50 + (s[3] - 48);
        c = (s[4] - 48) * 50 + (s[5] - 48);
        //cin >> a >> b >> c;
        if (dis[a][b] <= c)
            ans.push_back(1);
        else
            ans.push_back(0);
    }
    for (int i = 0; i < ans.size(); ++i)
        cout << ans[i];
    cout << endl;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 5836kb

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: 5876kb

input:

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

output:

010101

result:

ok single line: '010101'

Test #3:

score: -100
Wrong Answer
time: 57ms
memory: 5756kb

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:

010110001101001101111110111111100101111011110101000110110110101011010101111011111111111101001101011110111001110111100111111101011111101110001111100111101111110001100101110111111110100101111011001110101101111111110111100011111111101111011001010100111110110111110111111100111111101110111111011001111110...

result:

wrong answer 1st lines differ - expected: '011110001101101111111111111111...1111111111111111101111111111111', found: '010110001101001101111110111111...1101111101111100100011111111111'