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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 DetectionForever_Young#TL 12ms107120kbC++142.6kb2024-10-26 15:18:222024-10-26 15:18:22

Judging History

你现在查看的是最新测评结果

  • [2024-10-26 15:18:22]
  • 评测
  • 测评结果:TL
  • 用时:12ms
  • 内存:107120kb
  • [2024-10-26 15:18:22]
  • 提交

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 ...

output:


result: