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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
#132314#5668. Cell Nuclei Detectionwillow#AC ✓2056ms69340kbC++142.9kb2023-07-29 15:03:032023-07-29 15:03:04

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.
  • [2023-07-29 15:03:04]
  • 评测
  • 测评结果:AC
  • 用时:2056ms
  • 内存:69340kb
  • [2023-07-29 15:03:03]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int T, n, m;
namespace HK {
	int n1, n2, m, cx[maxn], cy[maxn], dx[maxn], dy[maxn], vis[maxn], id[maxn];
	vector<pair<int, int> > e[maxn];
	int q[maxn], head, tail;

	void Init(int _n, int _m) {
		n1 = _n, n2 = _m, m = 0;
		for(int i = 1; i <= n1; ++ i)
			e[i].clear();
	}
	void Add(int u, int v) {
		e[u].push_back({v, ++ m});
	}
	int Match(int u) {
		for(auto edge : e[u]) {
			int v = edge.first;
			if(!vis[v] && dy[v] == dx[u] + 1) {
				vis[v] = 1;
				if(cy[v] == -1 || Match(cy[v])) {
					cx[u] = v;
					cy[v] = u;
					id[u] = edge.second;
					return 1;
				}
			}
		}
		return 0;
	}
	bool BFS() {
		for(int i = 1; i <= n1; ++ i)
			dx[i] = 0;
		for(int i = 1; i <= n2; ++ i)
			dy[i] = 0;
		head = 1, tail = 0;
		for(int i = 1; i <= n1; ++ i)
			if(cx[i] == -1) {
				q[++ tail] = i;
				dx[i] = 1;
			}
		bool flag = 0;
		while(head <= tail) {
			int u = q[head ++];
			for(auto edge : e[u]) {
				int v = edge.first;
				if(dy[v] == 0) {
					dy[v] = dx[u] + 1;
					if(cy[v] == -1) {
						flag = true;
					}
					else {
						dx[cy[v]] = dy[v] + 1;
						q[++ tail] = cy[v];
					}
				}
			}
		}
		return flag;
	}

	int Solve() {
		int ans = 0;
		for(int i = 1; i <= n1; ++ i)
			cx[i] = -1;
		for(int i = 1; i <= n2; ++ i)
			cy[i] = -1;
		while(BFS()) {
			for(int i = 1; i <= max(n1, n2); ++ i) {
				vis[i] = 0;
			}
			for(int i = 1; i <= n1; ++ i) {
				if(cx[i] == -1) {
					ans += Match(i);
				}
			}
		}
		return ans;
	}
}
int blx[maxn], brx[maxn], bly[maxn], bry[maxn];
int dlx[maxn], drx[maxn], dly[maxn], dry[maxn];
int area[maxn];
map<pair<int, int>, vector<int> > mpid;
int Check(int i, int j) {
	int lx = max(dlx[i], blx[j]);
	int rx = min(drx[i], brx[j]);
	int ly = max(dly[i], bly[j]);
	int ry = min(dry[i], bry[j]);
	if(lx >= rx)
		return 0;
	if(ly >= ry)
		return 0;
	return (rx - lx) * (ry - ly) * 2 >= area[j];
}
int main() {
	for(scanf("%d", &T); T --; ) {
		scanf("%d%d", &m, &n);
		HK :: Init(n, m);
		for(int i = 1; i <= m; ++ i) {
			scanf("%d%d%d%d", &blx[i], &bly[i], &brx[i], &bry[i]);
			if(blx[i] > brx[i])
				swap(blx[i], brx[i]);
			if(bly[i] > bry[i])
				swap(bly[i], bry[i]);
			mpid[{blx[i], bly[i]}].push_back(i);
			area[i] = (brx[i] - blx[i]) * (bry[i] - bly[i]);
		}
		for(int i = 1; i <= n; ++ i) {
			scanf("%d%d%d%d", &dlx[i], &dly[i], &drx[i], &dry[i]);
			if(dlx[i] > drx[i])
				swap(dlx[i], drx[i]);
			if(dly[i] > dry[i])
				swap(dly[i], dry[i]);
		}
		for(int i = 1; i <= n; ++ i) {
			for(int x = max(0, dlx[i] - 4); x <= drx[i]; ++ x) {
				for(int y = max(0, dly[i] - 4); y <= dry[i]; ++ y) {
					if(mpid.find({x, y}) == mpid.end())
						continue;
					for(auto j : mpid[{x, y}])
						if(Check(i, j))
							HK :: Add(i, j);
				}
			}
		}
		printf("%d\n", HK :: Solve());
	}
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 10056kb

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: 2ms
memory: 10056kb

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: 1109ms
memory: 29272kb

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: 1073ms
memory: 39072kb

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: 497ms
memory: 21928kb

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: 2056ms
memory: 69340kb

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: 3ms
memory: 10052kb

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'