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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#89361#5668. Cell Nuclei DetectionFUCKUCUPAC ✓602ms113520kbC++141.9kb2023-03-19 21:11:302023-03-19 21:11:33

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.
  • [2023-03-19 21:11:33]
  • 评测
  • 测评结果:AC
  • 用时:602ms
  • 内存:113520kb
  • [2023-03-19 21:11:30]
  • 提交

answer

#include<bits/stdc++.h>
#define ri register int
#define fp(i, a, b) for(ri i = (a), ed = (b); i <= ed; ++i)
#define fb(i, a, b) for(ri i = (a), ed = (b); i >= ed; --i)
#define go(u, i) for(int i = head[u]; i; i = e[i].nxt)
using namespace std;

const int maxn = 100010, maxm = 2010;

int rd(){
	int f = 1, x = 0; char ch = getchar();
	while(ch > '9' || ch < '0'){ 	if(ch == '-')f = -1; ch = getchar(); }
	while(ch <= '9' && ch >= '0'){ x = (x << 1) + (x << 3) + (ch ^ 48);  ch = getchar(); }
	return x * f;
}

int T, m, n, lx[maxn], ly[maxn], rx[maxn], ry[maxn], mx, my;
int mat[maxn], vis[maxn], now;
vector<int> vec[maxm][maxm];
struct edge{int to, nxt;}e[maxn*100];
int head[maxn], k;
inline void add(int u, int v){e[++k] = (edge){v, head[u]}, head[u] = k;}
bool dfs(int u){
	vis[u] = now;
	go(u, i){
		if(!mat[e[i].to])return mat[u] = e[i].to, mat[e[i].to] = u, 1;
		if(vis[mat[e[i].to]] != now && dfs(mat[e[i].to]))return mat[u] = e[i].to, mat[e[i].to] = u, 1;
	}
	return 0;
}
int main(){
	T = rd();
	while(T--){
		m = rd(), n = rd();
		fp(i, 1, m+n)lx[i] = rd(), ly[i] = rd(), rx[i] = rd(), ry[i] = rd();
		mx = my = 0;
		fp(i, 1, m+n)mx = max(mx, rx[i]+4), my = max(my, ry[i]+4);
		fp(i, 0, mx)fp(j, 0, my)vec[i][j].clear();
		fp(i, m+1, m+n)vec[rx[i]][ry[i]].push_back(i);
		fp(i, 1, m+n)mat[i] = head[i] = vis[i] = 0;
		k = now = 0;
		fp(i, 1, m){
			fp(x, lx[i], rx[i]+4)fp(y, ly[i], ry[i]+4){
				for(auto &d : vec[x][y]){
					int nlx = max(lx[i], lx[d]), nly = max(ly[i], ly[d]);
					int nrx = min(rx[i], rx[d]), nry = min(ry[i], ry[d]);
					if(nlx >= nrx || nly >= nry)continue;
					if((nrx-nlx)*(nry-nly)*2 >= (rx[i]-lx[i])*(ry[i]-ly[i]))add(i, d);
				}
			}
		}
		int ans = 0;
		while(1){
			int cnt = 0;
			++now;
			fp(i, 1, m)if(!mat[i] && dfs(i))++cnt;
			if(!cnt)break;
			ans += cnt;
		}
		printf("%d\n", ans);
	}
}

Details

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Test #1:

score: 100
Accepted
time: 20ms
memory: 98256kb

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: 19ms
memory: 98304kb

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: 580ms
memory: 113520kb

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: 128ms
memory: 112756kb

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: 191ms
memory: 108024kb

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: 602ms
memory: 110028kb

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: 29ms
memory: 98460kb

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'