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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 |
---|---|---|---|---|---|---|---|---|---|
#678541 | #5668. Cell Nuclei Detection | Forever_Young# | TL | 12ms | 107120kb | C++14 | 2.6kb | 2024-10-26 15:18:22 | 2024-10-26 15:18:22 |
Judging History
answer
#include<bits/stdc++.h>
#define y1 yy
using namespace std;
const int N=50050;
const int M=500500;
const int inf=1000000000;
int t,x1[2*N],x2[2*N],y1[2*N],y2[2*N],S,T,res,n,m;
vector<int> mtr[2020][2020];
struct EDGE { int adj, w, next; } edge[M];
int top, gh[N], nrl[N];
void addedge(int x, int y, int w) {
//printf("%d %d %d\n",x,y,w);
edge[++top].adj = y;
edge[top].w = w;
edge[top].next = gh[x];
gh[x] = top;
edge[++top].adj = x;
edge[top].w = 0;
edge[top].next = gh[y];
gh[y] = top;
}
int dist[N], q[N];
int bfs() {
memset(dist, 0, sizeof(dist));
q[1] = S; int head = 0, tail = 1; dist[S] = 1;
while (head != tail) {
int x = q[++head];
for (int p=gh[x]; p; p=edge[p].next)
if (edge[p].w && !dist[edge[p].adj]) {
dist[edge[p].adj] = dist[x] + 1;
q[++tail] = edge[p].adj;
}
}
return dist[T];
}
int dinic(int x, int delta) {
if (x==T) return delta;
for (int& p=nrl[x]; p && delta; p=edge[p].next)
if (edge[p].w && dist[x]+1 == dist[edge[p].adj]) {
int dd = dinic(edge[p].adj, min(delta, edge[p].w));
if (!dd) continue;
edge[p].w -= dd;
edge[p^1].w += dd;
return dd;
}
return 0;
}
int work() {
int ans = 0;
while (bfs()) {
memcpy(nrl, gh, sizeof(gh));
int t; while (t = dinic(S, inf)) ans += t;
}
return ans;
}
int uni(int x,int y){
int a,b,c,d;
a=max(x1[x],x1[y]);
b=max(y1[x],y1[y]);
c=min(x2[x],x2[y]);
d=min(y2[x],y2[y]);
return max(0,c-a)*max(0,d-b);
}
int main(){
//freopen("I.in","r",stdin);
scanf("%d",&t);
while (t--){
scanf("%d%d",&m,&n);
S=0;
T=m+n+1;
for (int i=1;i<=m;i++){
scanf("%d%d%d%d",&x1[i],&y1[i],&x2[i],&y2[i]);
addedge(S,i,1);
}
for (int i=1;i<=n;i++){
scanf("%d%d%d%d",&x1[m+i],&y1[m+i],&x2[m+i],&y2[m+i]);
mtr[x1[m+i]][y1[m+i]].push_back(m+i);
addedge(i+m,T,1);
}
for (int i=1;i<=m;i++)
for (int j=max(0,x1[i]-3);j<=x1[i]+3;j++)
for (int k=max(0,y1[i]-3);k<=y1[i]+3;k++)
for (int h=0;h<mtr[j][k].size();h++){
if (2*uni(i,mtr[j][k][h])>=(x2[i]-x1[i])*(y2[i]-y1[i])) addedge(i,mtr[j][k][h],1);
}
res=work();
printf("%d\n",res);
top=0;
memset(gh,0,sizeof(gh));
memset(edge,0,sizeof(edge));
for (int i=1;i<=n;i++) mtr[x1[m+i]][y1[m+i]].clear();
}
}
详细
Test #1:
score: 100
Accepted
time: 12ms
memory: 107120kb
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: 8ms
memory: 107088kb
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: -100
Time Limit Exceeded
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 ...