【鏈接】 我是鏈接,點我呀:)
【題意】

在這裏輸入題意

【題解】

添加兩個y坐標0和h
然後從這n+2個y坐標中任選兩個坐標,作為矩形的上下界。

枚舉所有情況就好。

【代碼】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
 

#define all(x) x.begin(),x.end()
#define pb push_back
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1
using namespace std;

const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};
const int N = 100;

int n,w,h,px,py,anslen;
pair<int,int> a[N+10];

bool cmp(pair<int
 
,int> x,pair<int,int> y){
    return x.first<y.first;
}

void solve(){
    anslen = -1;
    cin >> n >> w >> h;
    for (int i = 1;i <= n;i++)
        cin >> a[i].first>>a[i].second;
    sort(a+1,a+1+n);

    a[0].second = 0;a[n+1].second = h;
    a[n+1].first = w;

    for
 
 (int i = 0;i <= n+1;i++){
        for (int j = 0;j <= n+1;j++){
            int upy = max(a[i].second,a[j].second),downy = min(a[i].second,a[j].second);
            int len = upy-downy;
            int prex = 0;
            for (int k = 1;k <= n+1;k++){
                if (k==n+1||(downy<a[k].second && a[k].second<upy)){
                    int templen = min(len,a[k].first-prex);
                    if (templen>0){
                        if (anslen==-1 || templen>anslen){
                                anslen = templen;
                                px = prex,py = downy;
                        }
                    }
                    prex = a[k].first;
                }
            }
        }
    }
    cout<<px<<' '<<py<<' '<<anslen<<endl;
}

int main(){
    #ifdef LOCAL_DEFINE
        freopen("rush_in.txt", "r", stdin);
    #endif
    ios::sync_with_stdio(0),cin.tie(0);
    int T;
    cin >> T;
    int kase = 0;
    while(T--){
        if (kase==0)
            kase++;
        else
            cout<<endl;
        solve();
    }
    return 0;
}

【習題 8-19 UVA-1312】Cricket Field