2015.10.20 2014--- 广州

前两天练的了，

做了过的人最多的三道后

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wtw发现第一题据说是西工大说的那个暴力出奇迹的题

第四题刚推出轨迹是个圆，，syh发现 ，，， 求的是圆和多边形的面积交---

于是弃疗又开了另一场比赛----

Dogs‘ Candies

The E-pang Palace

从30个点里面选择8个点组成两个矩形，使得它们的面积和最大

枚举每个矩形的左下，右上这两个点---

虽然意识到了有大矩形包含小矩形的情况----

这种包含的情况的时候，只算外面那个大矩形的面积----（sb地还把两个加起来----wa了一天---好忧桑---）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

typedef pair<int,int> pii;
int n;
int g[505][505];

struct node{
    int x,y;
}p[55];
int res;
int xx1,yy1,xx2,yy2,xx3,yy3,xx4,yy4;
int x1,y1,x2,y2,x3,y3,x4,y4;

int check(){
    if(x3 < x1 && y3 < y1 && x4 > x2 && y4 > y2) return 2;
    if(x3 > x2) return 1;
    if(y3 > y2) return 1;
    if(x4 < x1) return 1;
    if(y4 < y1) return 1;
    if(x3 == x2 && (y3 > y2 || yy3 < y1)) return 1;
    if(y3 == y2 && (x3 > x2 || xx4 < x1)) return 1;
    if(x4 == x1 && (yy4 > yy1 || y4 < y1)) return 1;
    if(y4 == y1 && (x4 < x1 || xx3 > xx2)) return 1;
    return 0;
}

int ok(){
    vector<pii> cc;
    int num = 0;
    if(g[xx1][yy1]){
        num++;
        g[xx1][yy1] = 0;
        cc.push_back(make_pair(xx1,yy1));
    }
    if(g[xx2][yy2]){
        num++;
        g[xx2][yy2] = 0;
        cc.push_back(make_pair(xx2,yy2));
    }
    if(g[xx3][yy3]){
        num++;
        g[xx3][yy3] = 0;
        cc.push_back(make_pair(xx3,yy3));
    }
    if(g[xx4][yy4]){
        num++;
        g[xx4][yy4] = 0;
        cc.push_back(make_pair(xx4,yy4));
    }
    for(int i = 0;i < cc.size();i++){
        int fi = cc[i].first,se = cc[i].second;
        g[fi][se] = 1;
    }
    return num == 4;
}

void work(int n1,int n2,int n3,int n4){
    int q[5];
    q[1] = n1;q[2] = n2;q[3] = n3;q[4] = n4;
    do{
        x1 = p[q[1]].x; y1 = p[q[1]].y;
        x2 = p[q[2]].x; y2 = p[q[2]].y;
        x3 = p[q[3]].x; y3 = p[q[3]].y;
        x4 = p[q[4]].x; y4 = p[q[4]].y;
        if(x1 >= x2 || y1 >= y2) continue;
        if(x3 >= x4 || y3 >= y4) continue;

        int a = x2-x1;
        int b = y2-y1;
        int c = x4-x3;
        int d = y4-y3;

        xx1 = x1;yy1 = y1 + b;
        xx2 = x1 + a;yy2 = y1;
        xx3 = x3;yy3 = y3 + d;
        xx4 = x3 + c;yy4 = y3;
        if(!ok()) continue;
        if(check() == 1) {
            res = max(res,a*b + c*d);
            //for(int i = 1;i <= 4;i++) printf("q[%d] = %d\n",i,q[i]);
        }
        if(check() == 2){
            res = max(res,max(a*b,c*d));
        }
    }while(next_permutation(q+1,q+5));
}

void solve(){
    if(n <= 7){
        puts("imp");
        return;
    }

    res = -1;
    for(int i = 1;i <= n;i++){
        for(int j = i+1;j <= n;j++){
            for(int k = j+1;k <= n;k++){
                for(int z = k+1;z <= n;z++){
                  // printf("i = %d  j = %d  k = %d  z = %d\n",i,j,k,z);
                    work(i,j,k,z);
                }
            }
        }
    }
    if(res == -1) puts("imp");
    else printf("%d\n",res);
}

int main() {
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d",&n) != EOF && n){
        memset(g,0,sizeof(g));
        for(int i = 1;i <= n;i++){
            scanf("%d %d",&p[i].x,&p[i].y);
            g[p[i].x][p[i].y] = 1;
        }
        solve();
    }
    return 0;
}

Yong Zheng‘s Death

Signal Interference

Song Jiang‘s rank list

Train Scheduling

Squared Frequency

Highway

Little Zu Chongzhi‘s Triangles

Yue Fei‘s Battle

How Many Maos Does the Guanxi Worth