HDU 5023 A Corrupt Mayor's Performance Art （计算几何、血坑）

题意:

N<=30点,求2个不cross,不touch的矩形的最大面积和,拼不出2个矩形输出“imp”

分析:

暴力对角线2个点,然后另外2个点可以算,得到第一个矩形
再暴力2个点,再算出2个点,得到第二个矩形
第二个矩形不能和第一个touch,cross,其实就是这个4个点不在前面那个矩形范围内

血坑:

大的完全包含小的矩形的情况是合法的。。。。。。

代码:

////  Created by TaoSama on 2015-10-23//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;struct Point {    int x, y;    Point(int x = 0, int y = 0): x(x), y(y) {}    void input() {        scanf("%d%d", &x, &y);    }    void print() {        cout << x << ' ' << y << endl;    }} A[35];bool vis[205][205];int l, r, up, down;bool onLine(Point& a, Point& b) {    return a.x == b.x || a.y == b.y;}bool isLegal(Point& a) {    if(a.x >= l && a.x <= r && a.y >= down && a.y <= up) return 0;    return 1;}bool isIn(Point& a) {    return (a.x > l && a.x < r && a.y > down && a.y < up);}int getRectArea(Point a, Point b, Point& c, Point& d) {    if(a.x > b.x) swap(a, b);    c = Point(a.x, b.y), d = Point(b.x, a.y);    if(vis[c.x][c.y] && vis[d.x][d.y])        return abs(a.x - b.x) * abs(a.y - b.y);    return -INF;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif//    ios_base::sync_with_stdio(0);    while(scanf("%d", &n) == 1 && n) {        memset(vis, false, sizeof vis);        for(int i = 1; i <= n; ++i) {            A[i].input();            vis[A[i].x][A[i].y] = true;        }        int ans = -INF;        Point C, D;        for(int a = 1; a <= n; ++a) {            for(int b = 1; b <= n; ++b) {                if(onLine(A[a], A[b])) continue;                int tmp = getRectArea(A[a], A[b], C, D);                if(tmp == -INF) continue;                l = min(A[a].x, A[b].x), r = max(A[a].x, A[b].x);                down = min(A[a].y, A[b].y), up = max(A[a].y, A[b].y);                for(int c = 1; c <= n; ++c) {                    for(int d = 1; d <= n; ++d) {                        if(onLine(A[c], A[d])) continue;                        int tmp2 = getRectArea(A[c], A[d], C, D);                        if(tmp2 == -INF || tmp < tmp2) continue;                        if(isIn(A[c]) && isIn(A[d]) && isIn(C) && isIn(D)) {                            ans = max(ans, tmp);                        }//                        if(tmp == 3){//                            A[a].print();//                            A[b].print();//                            A[c].print();//                            A[d].print();//                            cout << "max" << endl;//                        }                        if(!isLegal(A[c]) || !isLegal(A[d]) || !isLegal(C) || !isLegal(D))                            continue;                        ans = max(ans, tmp + tmp2);//                        if(tmp + tmp2 == 3){//                            A[a].print();//                            A[b].print();//                            A[c].print();//                            A[d].print();//                            cout << endl;//                        }                    }                }            }        }        if(ans != -INF) printf("%d\n", ans);        else puts("imp");    }    return 0;}