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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
#518394#1819. Cleaning RobotAA_Surely#WA 140ms159928kbC++232.9kb2024-08-13 20:09:262024-08-13 20:09:26

Judging History

你现在查看的是最新测评结果

  • [2024-08-13 20:09:26]
  • 评测
  • 测评结果:WA
  • 用时:140ms
  • 内存:159928kb
  • [2024-08-13 20:09:26]
  • 提交

answer

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define FOR(i, x, n) for(int i = x; i < n; i++)
#define F0R(i, n) FOR(i, 0, n)
#define ROF(i, x, n) for(int i = n - 1; i >= x; i--)
#define R0F(i, n) ROF(i, 0, n)

#define WTF cout << "WTF" << endl

#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define F first
#define S second
#define PB push_back
#define EP emplace_back

#define ALL(x) x.begin(), x.end()
#define RALL(x) x.rbegin(), x.rend()

using namespace std;

typedef long long LL;
typedef long double LD;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

typedef vector<int> VI;
typedef vector<LL> VLL;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;

typedef complex<LD> Point;

const int N = 1e6 + 7;
const int INF = 1e9 + 7;
const int LOG = 22;
const int A = 26;
const int SQ = 320;
const int MOD = 1e9 + 7;

LL n, m, k;
VLL ns[N], ps[N], can[N], seen[N];

int xm[4] = {1, 0, 0,-1};
int ym[4] = {0,-1, 1, 0};

LL getPs(int i1, int j1, int i2, int j2) {
    LL ret = ps[i2][j2];
    if (i1) ret -= ps[i1 - 1][j2];
    if (j1) ret -= ps[i2][j1 - 1];
    if (i1 && j1) ret += ps[i1 - 1][j1 - 1];

    return ret;
}

bool canGo(int i, int j, int l) {
    //cout << " i j " <<i << ' ' << j << ' ' << l << endl;
    if (i < 0 || j < 0) return 0;
    if (i + l - 1 >= n || j + l - 1 >= m) return 0;
    if (getPs(i, j, i + l - 1, j + l - 1)) return 0;
    return 1;
}

void dfs(int x, int y, int l) {
    seen[x][y] = 1;
    can[x][y] = 1;

    F0R(i, 4)
        if (canGo(x + xm[i], y + ym[i], l) && !seen[x + xm[i]][y + ym[i]]) 
            dfs(x + xm[i], y + ym[i], l);
    
    return;
}

LL getCan(int i1, int j1, int i2, int j2) {
    LL ret = can[i2][j2];
    if (i1 > 0) ret -= can[i1 - 1][j2];
    if (j1 > 0) ret -= can[i2][j1 - 1];
    if (i1 > 0 && j1 > 0) ret += can[i1 - 1][j1 - 1];

    return ret;
}

bool check(int x) {
    F0R(i, n) F0R(j, m) can[i][j] = seen[i][j] = 0;

    F0R(i, n) F0R(j, m) if (canGo(i, j, x)) {
        dfs(i, j, x);
        F0R(ii, n) 
            FOR(jj, 1, m) can[ii][jj] += can[ii][jj - 1];
        FOR(ii, 1, n) F0R(jj, m)
            can[ii][jj] += can[ii - 1][jj];

        F0R(ii, n) F0R(jj, m) {
            if (!getCan(ii - x + 1, jj - x + 1, ii, jj) && !ns[ii][jj]) {
                return 0;
            }
        }
        return 1;
    }

    return 1;
}

int main() {
    IOS;
    
    cin >> n >> m >> k;
    F0R(i, n) {
        ns[i].resize(m);
        ps[i].resize(m);
        seen[i].resize(m);
        can[i].resize(m);
    }

    F0R(i, k) {
        int a, b;
        cin >> a >> b;
        ns[--a][--b] = 1;
    }

    F0R(i, n) {
        ps[i][0] = ns[i][0];
        FOR(j, 1, m) ps[i][j] = ps[i][j - 1] + ns[i][j];
    }

    FOR(i, 1, n) F0R(j, m)
        ps[i][j] += ps[i - 1][j];

    int l = 1, r = min(n, m);

    while(l < r) {
        int mid = (l + r) >> 1;
        if (check(mid)) l = mid + 1;
        else r = mid;
    }

    cout << (l - 1 <= 0 ? -1 : l - 1) << endl;
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 3568kb

input:

10 7 1
8 3

output:

2

result:

ok answer is '2'

Test #2:

score: -100
Wrong Answer
time: 140ms
memory: 159928kb

input:

2236 2236 2214
28 1255
389 2175
730 592
1360 977
1225 752
1403 1798
1518 1381
147 745
659 249
951 1475
1826 1951
691 1033
81 1458
1487 1946
2106 1395
1995 629
470 891
1902 822
2210 2001
441 2130
1198 1539
2027 1101
215 1149
205 420
379 2104
308 1225
859 109
1417 2078
1764 376
1772 5
335 1113
917 118...

output:

2235

result:

wrong answer expected '1', found '2235'