2019HDU多校八 6665 Calabash and Landlord（离散化BFS）

hdu6665

题意：

求两个矩形在二维平面内组成的封闭区域的个数

思路：

暴力讨论

赛中队友想暴力搞一下就被我否决了（情况太多有这时间还不如去开其他题？）

考虑对矩形离散化到一个$15\ast 15$的区域内，矩形的边界用特殊的符号表示，然后对这个区域进行$bfs$就完事了。

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9 + 7;
using namespace std;
struct Node{
    ll a, b;
    bool operator < (const Node x) const{
        return b > x.b;
    }
}p[maxn];
vector<int> xx, yy;
int vis[15][15];
struct cp{
    int x, y;
};
int to[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
void bfs(int x, int y, int id){
    queue<cp> q;
    while(!q.empty()) q.pop();
    vis[x][y] = id;
    cp a, b;
    a.x = x, a.y = y;
    q.push(a);
    while(!q.empty()){
        a = q.front();
        q.pop();
        for(int i = 0; i < 4; i++){
            b.x = a.x + to[i][0];
            b.y = a.y + to[i][1];
            if(b.x < 0 || b.y < 0 || b.x >= 15 || b.y >= 15) continue;
            if(vis[b.x][b.y] != 0) continue;
            vis[b.x][b.y] = id;
            q.push(b);
        }
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
       int x[5], y[5];
       xx.clear(), yy.clear();
       scanf("%d%d%d%d", &x[1], &y[1], &x[2], &y[2]);
       scanf("%d%d%d%d", &x[3], &y[3], &x[4], &y[4]);
       for(int i = 1; i <= 4; i++){
            xx.push_back(x[i]);
            yy.push_back(y[i]);
       }
       sort(xx.begin(), xx.end()), sort(yy.begin(), yy.end());
       xx.erase(unique(xx.begin(), xx.end()), xx.end());
       yy.erase(unique(yy.begin(), yy.end()), yy.end());
       for(int i = 1; i <= 4; i++){
            x[i] = lower_bound(xx.begin(), xx.end(), x[i]) - xx.begin() + 1;
            x[i] += (x[i] - 1);
       }
       for(int i = 1; i <= 4; i++){
            y[i] = lower_bound(yy.begin(), yy.end(), y[i]) - yy.begin() + 1;
            y[i] += (y[i] - 1);
       }

       memset(vis, 0, sizeof(vis));
       for(int i = x[1]; i <= x[2]; i++){
            vis[i][y[1]] = -1;
            vis[i][y[2]] = -1;
       }
       for(int i = x[3]; i <= x[4]; i++){
            vis[i][y[3]] = -1;
            vis[i][y[4]] = -1;
       }
       for(int i = y[1]; i <= y[2]; i++){
            vis[x[1]][i] = -1;
            vis[x[2]][i] = -1;
       }
       for(int i = y[3]; i <= y[4]; i++){
            vis[x[3]][i] = -1;
            vis[x[4]][i] = -1;
       }
       int ans = 0;
       for(int i = 0; i < 15; i++){
            for(int j = 0; j < 15; j++){
                if(vis[i][j] != -1 && vis[i][j] == 0){
                    ans++;
                    bfs(i, j, ans);
                }
            }
       }
       printf("%d\n", ans);
    }
    return 0;
}