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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
#104096#5668. Cell Nuclei DetectiontarjenAC ✓1287ms162020kbC++174.1kb2023-05-08 16:28:352023-05-08 16:28:38

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.
  • [2023-05-08 16:28:38]
  • 评测
  • 测评结果:AC
  • 用时:1287ms
  • 内存:162020kb
  • [2023-05-08 16:28:35]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = 20 * (1e5);
class Maxflow
{
private:
    int nedge = 1, p[2 * M], nex[2 * M], head[N], c[2 * M], cur[2 * M];
    int dist[2 * N];
    int S, T;
    void Addedge(int a, int b, int v)
    {
        p[++nedge] = b;
        nex[nedge] = head[a];
        head[a] = nedge;
        c[nedge] = v;
    }
    bool bfs()
    {
        queue<int> q;
        for (int i = S; i <= T; i++)
            dist[i] = -1;
        dist[S] = 0;
        q.push(S);
        while (!q.empty())
        {
            int now = q.front();
            q.pop();
            for (int k = head[now]; k; k = nex[k])
                if (dist[p[k]] == -1 && c[k] > 0)
                {
                    dist[p[k]] = dist[now] + 1;
                    q.push(p[k]);
                }
        }
        return dist[T] > -1;
    }
    int dfs(int x, int low)
    {
        if (x == T)
            return low;
        if (low == 0)
            return 0;
        int used = 0;
        for (int &k = cur[x]; k; k = nex[k])
            if (dist[p[k]] == dist[x] + 1 && c[k] > 0)
            {
                int a = dfs(p[k], min(c[k], low - used));
                c[k] -= a;
                c[k ^ 1] += a;
                used += a;
                if (low == used)
                    break;
            }
        if (used == 0)
            dist[x] = -1;
        return used;
    }

public:
    void givest(int s, int t)
    {
        S = s;
        T = t;
    }
    void init()
    {
        memset(head, 0, sizeof(head));
        nedge = 1;
    }
    void addedge(int a, int b, int v)
    {
        //cout << a << " " << b << " "<< v <<endl;
        Addedge(a, b, v);
        Addedge(b, a, 0);
    }
    int dinic()
    {
        int flow = 0;
        while (bfs())
        {
            for (int i = S; i <= T; i++)
                cur[i] = head[i];
            flow += dfs(S, 1e9);
        }
        return flow;
    }
};
Maxflow e;
vector<int> g[2001][2001];
struct node
{
    int x1, y1, x2, y2;
} a[50010], b[50010];
int f[2001][2001], cnt;
bool check(node x, node y) // y 要占 x 的一半
{
    cnt++;
    int sum = 0, ss = 0;
    for (int i = x.x1; i < x.x2; i++)
    {
        for (int j = x.y1; j < x.y2; j++)
        {
            f[i][j] = cnt;
            sum++;
        }
    }
    for (int i = y.x1; i < y.x2; i++)
    {
        for (int j = y.y1; j < y.y2; j++)
        {
            if (f[i][j] == cnt)
                ss++;
        }
    }
    if (ss * 2 >= sum)
        return 1;
    return 0;
}
void work()
{
    e.init();
    int m, n;
    int s = 0, t;
    scanf("%d%d", &m, &n); // m:true n:detect
    t = m + n + 1;
    e.givest(s,t);
    for (int i =1;i<=m;i++)
    {
        e.addedge(s,i,1);
    }
    for (int i =1;i<=n;i++)
    {
        e.addedge(i+m,t,1);
    }

    for (int i = 0; i <= 2000; i++)
        for (int j = 0; j <= 2000; j++)
            g[i][j].clear();
    for (int i = 1, x1, y1, x2, y2; i <= m; i++)
    {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        a[i] = node{x1, y1, x2, y2};
        for (int x = x1; x < x2; x++)
            for (int y = y1; y < y2; y++)
            {
                g[x][y].push_back(i);
            }
    }
    for (int i = 1, x1, y1, x2, y2; i <= n; i++)
    {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        b[i] = node{x1, y1, x2, y2};
        vector<int> q;
        for (int x = x1; x < x2; x++)
            for (int y = y1; y < y2; y++)
                for (auto z : g[x][y])
                {
                    q.push_back(z);
                }
        sort(q.begin(), q.end());
        for (int j = 0; j < (int)q.size(); j++)
        {
            if (j == 0 || q[j] != q[j - 1])
            {
                if (check(a[q[j]], b[i]))
                {
                    e.addedge(q[j], i + m, 1);
                }
            }
        }
        
    }
    cout << e.dinic() << endl;
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        work();
    }
}

详细

Test #1:

score: 100
Accepted
time: 35ms
memory: 107996kb

input:

3
2 2
1 1 3 3
3 3 5 5
2 2 4 4
4 4 6 6
2 3
1 1 3 3
3 3 5 5
1 3 3 5
2 1 4 5
3 1 5 3
3 3
1 1 2 2
2 2 3 3
3 3 4 4
1 1 3 3
2 2 4 4
3 3 5 5

output:

0
1
3

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 31ms
memory: 109988kb

input:

3
2 2
1 1 3 3
3 3 5 5
2 2 4 4
4 4 6 6
2 3
1 1 3 3
3 3 5 5
1 3 3 5
2 1 4 5
3 1 5 3
3 3
1 1 2 2
2 2 3 3
3 3 4 4
1 1 3 3
2 2 4 4
3 3 5 5

output:

0
1
3

result:

ok 3 lines

Test #3:

score: 0
Accepted
time: 605ms
memory: 156548kb

input:

5
50000 50000
0 0 4 4
4 0 8 4
8 0 12 4
12 0 16 4
16 0 20 4
20 0 24 4
24 0 28 4
28 0 32 4
32 0 36 4
36 0 40 4
40 0 44 4
44 0 48 4
48 0 52 4
52 0 56 4
56 0 60 4
60 0 64 4
64 0 68 4
68 0 72 4
72 0 76 4
76 0 80 4
80 0 84 4
84 0 88 4
88 0 92 4
92 0 96 4
96 0 100 4
100 0 104 4
104 0 108 4
108 0 112 4
112 ...

output:

50000
50000
0
50000
3150

result:

ok 5 lines

Test #4:

score: 0
Accepted
time: 421ms
memory: 162020kb

input:

5
50000 50000
0 0 1 1
1 0 2 1
2 0 3 1
3 0 4 1
4 0 5 1
5 0 6 1
6 0 7 1
7 0 8 1
8 0 9 1
9 0 10 1
10 0 11 1
11 0 12 1
12 0 13 1
13 0 14 1
14 0 15 1
15 0 16 1
16 0 17 1
17 0 18 1
18 0 19 1
19 0 20 1
20 0 21 1
21 0 22 1
22 0 23 1
23 0 24 1
24 0 25 1
25 0 26 1
26 0 27 1
27 0 28 1
28 0 29 1
29 0 30 1
30 0 ...

output:

50000
25050
12500
16000
8000

result:

ok 5 lines

Test #5:

score: 0
Accepted
time: 389ms
memory: 140308kb

input:

5
50000 50000
0 0 2 4
4 0 7 1
8 0 10 1
12 0 15 3
16 0 19 1
20 0 22 2
24 0 26 4
28 0 30 4
32 0 36 3
36 0 40 1
40 0 44 1
44 0 47 2
48 0 49 3
52 0 54 1
56 0 59 4
60 0 64 3
64 0 68 3
68 0 70 1
72 0 76 4
76 0 80 3
80 0 84 4
84 0 87 2
88 0 90 1
92 0 94 4
96 0 98 1
100 0 104 1
104 0 107 2
108 0 110 4
112 0...

output:

10594
10779
10618
10381
10779

result:

ok 5 lines

Test #6:

score: 0
Accepted
time: 1287ms
memory: 126956kb

input:

5
50000 50000
0 0 4 4
1 0 5 4
2 0 6 4
3 0 7 4
4 0 8 4
5 0 9 4
6 0 10 4
7 0 11 4
8 0 12 4
9 0 13 4
10 0 14 4
11 0 15 4
12 0 16 4
13 0 17 4
14 0 18 4
15 0 19 4
16 0 20 4
17 0 21 4
18 0 22 4
19 0 23 4
20 0 24 4
21 0 25 4
22 0 26 4
23 0 27 4
24 0 28 4
25 0 29 4
26 0 30 4
27 0 31 4
28 0 32 4
29 0 33 4
30...

output:

50000
50000
50000
50000
49600

result:

ok 5 lines

Test #7:

score: 0
Accepted
time: 24ms
memory: 107956kb

input:

1
4 4
1 1 3 3
2 1 4 3
1 2 3 4
2 2 4 4
2 1 4 3
3 2 5 4
1 2 3 4
2 3 4 5

output:

3

result:

ok single line: '3'