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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
#88406#5668. Cell Nuclei DetectionCCSU_WineRE 0ms0kbC++203.2kb2023-03-16 10:20:542023-03-16 10:20:56

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.
  • [2023-03-16 10:20:56]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2023-03-16 10:20:54]
  • 提交

answer

#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2010, M = 5010, inf = 1e5, SIZ = 2000;

int n, m;
struct Box{
    int x1, y1, x2, y2, area;
    void input(){ cin >> x1 >> y1 >> x2 >> y2; }
    int get_id(){ return (x1 - 1) * SIZ + y1; } //返回左下角坐标对应编号
    void get_area(){
        area = (x2 - x1) * (y2 - y1);
    }
}tr[M], de[M];

struct edge{
    int to, nex, w;
}e[M * 200];
int tot, s, t, head[M * 2];
void add(int from, int to, int w){
    e[++ tot].w = w;
    e[tot].to = to;
    e[tot].nex = head[from];
    head[from] = tot;
}

int dep[M * 2];
bool bfs(){
    queue<int>q;
    q.push(s);
    for(int i = s; i <= t; i ++) dep[i] = 0;
    dep[s] = 1;
    while(!q.empty()){
        int u = q.front(); q.pop();
        for(int i = head[u]; i; i = e[i].nex){
            int v = e[i].to;
            if(e[i].w && !dep[v]){
                dep[v] = dep[u] + 1;
                if(v == t) return true;
                q.push(v);
            }
        }
    }
    return dep[t];
}

int dfs(int u, int inflow){
    if(u == t) return inflow;
    int outflow = 0;
    for(int i = head[u]; i && inflow; i = e[i].nex){
        int v = e[i].to;
        if(e[i].w && dep[v] == dep[u] + 1){
            int flow = dfs(v, min(e[i].w, inflow));
            e[i].w -= flow;
            e[i ^ 1].w += flow;
            inflow -= flow;
            outflow += flow;
        }
    }
    if(outflow == 0) dep[u] = 0;
    return outflow;
}

vector<int>g[N * N];//存坐标所有的矩形左上角编号
void init(){
    tot = 1;
    for(int i = 1; i <= m; i ++){
        g[tr[i].get_id()].clear();
    }
    for(int i = s; i <= t; i ++) head[i] = 0;
}


int overlap_area(Box &a, Box &b){//计算线段交
    int x = max(0, min(a.x2, b.x2) - max(a.x1, b.x1));
    int y = max(0, min(a.y2, b.y2) - max(a.y1, b.y1));
    return x * y;
}

void build_edge(int u){
    Box &A = de[u];
    for(int i = max(A.x1 - 2, 0); i <= A.x2 + 2; i ++){
        for(int j = max(A.y1 - 2, 0); j <= A.y2 + 2; j ++){
            int idx = (i - 1) * SIZ + j;
            for(int id : g[idx]){
                if(overlap_area(A, tr[id]) * 2 >= tr[id].area){
                    add(u, id + n, 1);
                    add(id + n, u, 0);
                }
            }
        }
    }
}

void solve(){
    init();
    cin >> m >> n;
    for(int i = 1; i <= m; i ++){//真实框
        tr[i].input();
        tr[i].get_area();
        g[tr[i].get_id()].push_back(i);
    }
    for(int i = 1; i <= n; i ++){// 搜索框
        de[i].input();
        de[i].get_area();
    }
    
    s = 0, t = n + m + 1;
    for(int i = 1; i <= n; i ++){
        add(s, i, 1);
        add(i, s, 0);
        build_edge(i);
    }
    for(int i = 1; i <= m; i ++){
        add(i + n, t, 1);
        add(t, i + n, 0);
    }

    int ans = 0;
    while(bfs()){
        ans += dfs(s, inf);
    }
    cout << ans << "\n";
}
int main(){
    ios::sync_with_stdio(false);
    cout.tie(NULL);
    int t;
    cin >> t;
    while(t --){
        solve();
    }
    return 0;
}

详细

Test #1:

score: 0
Runtime Error

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:


result: