标签：
题意：给了一些点，问组成两个不相交的矩形的面积和最大
分析：暴力枚举，先找出可以组成矩形的两点并保存起来(vis数组很好)，然后写个函数判断四个点是否在另一个矩形内部。当时没有保存矩形，用for来找矩形，结果写糊涂了忘记判断回形的情况。。。
/************************************************
* Author :Running_Time
* Created Time :2015/11/6 星期五 17:00:44
* File Name :B_2.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Point {
int x, y;
Point () {}
Point (int x, int y) : x (x), y (y) {}
bool operator < (const Point &r) const {
if (x == r.x) return y < r.y;
else return x < r.x;
}
}p[33];
struct Matrix {
Point a, b;
Matrix () {}
Matrix (Point a, Point b) : a (a), b (b) {}
};
vector<Matrix> mat;
bool vis[N][N];
int inside(Point p, Point a, Point b) {
if (p.x >= a.x && p.x <= b.x
&& p.y <= a.y && p.y >= b.y) {
if (p.x > a.x && p.x < b.x
&& p.y < a.y && p.y > b.y) return -1;
else return 1;
}
else return 0;
}
int area_mat(int i) {
return (mat[i].b.x - mat[i].a.x) * (mat[i].a.y - mat[i].b.y);
}
int judge(int i, int j) {
Point ic = Point (mat[i].a.x, mat[i].b.y),
id = Point (mat[i].b.x, mat[i].a.y);
int res1 = inside (mat[i].a, mat[j].a, mat[j].b);
int res2 = inside (mat[i].b, mat[j].a, mat[j].b);
int res3 = inside (ic, mat[j].a, mat[j].b);
int res4 = inside (id, mat[j].a, mat[j].b);
if (!res1 && !res2 && !res3 && !res4) return 0;
else if (res1 == -1 && res2 == -1 && res3 == -1 && res4 == -1) return -1;
else return 1;
}
int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
if (!n) break;
mat.clear ();
memset (vis, false, sizeof (vis));
for (int i=0; i<n; ++i) {
scanf ("%d%d", &p[i].x, &p[i].y);
vis[p[i].x][p[i].y] = true;
}
sort (p, p+n);
for (int i=0; i<n; ++i) {
int x1 = p[i].x, y1 = p[i].y;
for (int j=i+1; j<n; ++j) {
int x2 = p[j].x, y2 = p[j].y;
if (x1 >= x2 || y1 <= y2) continue;
if (!vis[x1][y2] || !vis[x2][y1]) continue;
mat.push_back (Matrix (Point (x1, y1), Point (x2, y2)));
}
}
int ans = 0;
for (int i=0; i<mat.size (); ++i) {
for (int j=i+1; j<mat.size (); ++j) {
int res1 = judge (i, j);
int res2 = judge (j, i);
if (!res1 && !res2) {
ans = max (ans, area_mat (i) + area_mat (j));
}
if (res1 == -1 || res2 == -1) {
ans = max (ans, max (area_mat (i), area_mat (j)));
}
}
}
if (ans == 0) puts ("imp");
else printf ("%d\n", ans);
}
//cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
return 0;
}
这个很挫的代码放在这留个念。。。
/************************************************
* Author :Running_Time
* Created Time :2015/10/14 星期三 14:59:33
* File Name :B.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
const int N = 33;
const int M = 210;
const int INF = 0x3f3f3f3f;
struct Point {
int x, y;
bool operator < (const Point &r) const {
if (x == r.x) return y < r.y;
else return x < r.x;
}
}p[N];
vector<int> vx[M], vy[M];
int cal(int la, int lb) {
if (la < 0) la = -la;
if (lb < 0) lb = -lb;
return la * lb;
}
int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
if (!n) break;
for (int i=0; i<=200; ++i) {
vx[i].clear (); vy[i].clear ();
}
for (int i=1; i<=n; ++i) {
scanf ("%d%d", &p[i].x, &p[i].y);
}
sort (p+1, p+1+n);
for (int i=1; i<=n; ++i) {
vx[p[i].x].push_back (i); vy[p[i].y].push_back (i);
}
int ans = 0;
bool flag = false;
for (int i=1; i<=n; ++i) { //i one
int x = p[i].x, y = p[i].y;
if (vx[x].size () <= 1 || vy[y].size () <= 1) continue;
for (int j=0; j<vx[x].size (); ++j) { //j two
int jj = vx[x][j];
if (jj == i || p[jj].y <= y) continue;
int jy = p[jj].y;
for (int k=0; k<vy[y].size (); ++k) { //k three
int kk = vy[y][k];
if (kk == i || p[kk].x <= x) continue;
int kx = p[kk].x;
if (vx[kx].size () <= 1) continue;
for (int l=0; l<vx[kx].size (); ++l) { //l four
int ll = vx[kx][l];
if (ll == kk || p[ll].y <= y) continue;
if (p[ll].y == jy) { //find the first rectangle
if (vy[jy].size () >= 4) {
for (int r=0; r<vy[jy].size (); ++r) {
int rr = vy[jy][r];
if (rr == ll || rr == jj) continue;
for (int r3=0; r3<vy[jy].size (); ++r3) {
int r4 = vy[jy][r3];
if (r4 == ll || r4 == jj || r4 == rr) continue;
int xx = p[rr].x, kkx = p[r4].x;
if (x <= xx && xx <= kx) continue;
if (x <= kkx && kkx <= kx) continue;
if (vx[xx].size () <= 1 || vx[kkx].size () <= 1) continue;
for (int o=0; o<vx[xx].size (); ++o) {
int oo = vx[xx][o];
if (oo == rr || p[oo].y <= jy) continue;
int jjy = p[oo].y;
for (int u=0; u<vx[kkx].size (); ++u) {
int uu = vx[kkx][u];
if (uu == r4 || p[uu].y != jjy) continue;
flag = true;
ans = max (ans, cal (kx - x, jy - y) + cal (kkx - xx, jjy - jy));
}
}
}
}
}
for (int yy=jy+1; yy<=200; ++yy) {
if (vy[yy].size () <= 1) continue;
for (int yi=0; yi<vy[yy].size (); ++yi) {
int ii = vy[yy][yi];
int xx = p[ii].x, yy = p[ii].y; //to find the second rectangle
for (int j3=0; j3<vx[xx].size (); ++j3) {
int j4 = vx[xx][j3];
if (j4 == ii || p[j4].y <= yy) continue;
int jjy = p[j4].y;
for (int k3=0; k3<vy[yy].size (); ++k3) {
int k4 = vy[yy][k3];
if (k4 == ii) continue;
int kkx = p[k4].x;
for (int l3=0; l3<vx[kkx].size (); ++l3) {
int l4 = vx[kkx][l3];
if (l4 == k4 || p[l4].y != jjy) continue;
flag = true;
ans = max (ans, cal (kx - x, jy - y) + cal (kkx - xx, jjy - yy));
}
}
}
}
}
if (vx[kx].size () >= 4) {
for (int r=0; r<vx[kx].size (); ++r) {
int rr = vx[kx][r];
if (rr == kk || rr == ll) continue;
for (int r3=0; r3<vy[jy].size (); ++r3) {
int r4 = vx[kx][r3];
if (r4 == kk || r4 == ll || r4 == rr) continue;
int yy = p[rr].y, jjy = p[r4].y;
if (yy > jjy) swap (yy, jjy);
if (y <= yy && yy <= jy) continue;
if (y <= jjy && jjy <= jy) continue;
if (vy[yy].size () <= 1 || vy[jjy].size () <= 1) continue;
for (int o=0; o<vy[yy].size (); ++o) {
int oo = vy[yy][o];
if (oo == rr || p[oo].x <= kx) continue;
int kkx = p[oo].x;
for (int u=0; u<vy[jjy].size (); ++u) {
int uu = vy[jjy][u];
if (uu == r4 || p[uu].x != kkx) continue;
flag = true;
ans = max (ans, cal (kx - x, jy - y) + cal (kkx - kx, jjy - jy));
}
}
}
}
}
for (int x5=kx+1; x5<=200; ++x5) {
if (vx[x5].size () <= 1) continue;
for (int xi=0; xi<vx[x5].size (); ++xi) {
int ii = vx[x5][xi];
int xx = p[ii].x, yy = p[ii].y; //to find the second rectangle
for (int j3=0; j3<vx[x5].size (); ++j3) {
int jj = vx[x5][j3];
if (jj == ii || p[jj].y <= yy) continue;
int jjy = p[jj].y;
if (vy[yy].size () <= 1) continue;
for (int k3=0; k3<vy[yy].size (); ++k3) {
int k4 = vy[yy][k3];
if (k4 == ii || p[k4].x <= xx) continue;
int kkx = p[k4].x;
for (int l3=0; l3<vx[kkx].size (); ++l3) {
int l4 = vx[kkx][l3];
if (l4 == k4 || p[l4].y != jjy) continue;
flag = true;
ans = max (ans, cal (kx - x, jy - y) + cal (kkx - xx, jjy - yy));
}
}
}
}
}
}
}
}
}
}
if (flag) {
printf ("%d\n", ans);
}
else puts ("imp");
}
return 0;
}
简单几何(判断矩形的位置) UVALive 7070 The E-pang Palace (14广州B)
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原文地址：http://www.cnblogs.com/Running-Time/p/4943280.html
