[USACO 6.2.2] Packing Rectangles

题目大意

　　给出4个矩形,求一个面积最小的矩形能够容纳着4个矩形,4个矩形不能互相叠起来.

题解

　　看USACO的图,它并没有给出全部情况.

　　还有两种情况.比较好想.这里不一一说明.

　　其实这题考察大家的缜密的思维以及是否拥有良好的代码习惯(因为有时候敲起来会很乱).

代码

/*
TASK:packrec
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct Matrix
{
    int h, w;

    bool operator < (const Matrix &b) const
    {
        return h < b.h || h == b.h && w < b.w;
    }
}ans[1024], minmat[4], a[4];

int anslen;
bool v[4];

void judge()
{
    if (ans[anslen].h > ans[anslen].w) swap(ans[anslen].h, ans[anslen].w);
    if (ans[anslen].h * ans[anslen].w < ans[0].h * ans[0].w)
    {
        ans[0] = ans[anslen];
        anslen = 1;
        return;
    }
    if (ans[anslen].h * ans[anslen].w == ans[0].h * ans[0].w) anslen++;
}

void dfs(int s)
{
    if (s == 4)
    {
        //1
        ans[anslen].h = max(minmat[0].h, max(minmat[1].h, max(minmat[2].h, minmat[3].h)));
        ans[anslen].w = minmat[0].w + minmat[1].w + minmat[2].w + minmat[3].w;
        judge();
        //2
        ans[anslen].h = max(minmat[0].h, max(minmat[1].h, minmat[2].h)) + minmat[3].h;
        ans[anslen].w = max(minmat[3].w, minmat[0].w + minmat[1].w + minmat[2].w);
        judge();
        //3
        ans[anslen].h = max(minmat[3].h, max(minmat[0].h, minmat[1].h) + minmat[2].h);
        ans[anslen].w = max(minmat[0].w + minmat[1].w, minmat[2].w) + minmat[3].w;
        judge();
        //4
        ans[anslen].h = max(minmat[0].h, max(minmat[1].h + minmat[2].h, minmat[3].h));
        ans[anslen].w = minmat[0].w + max(minmat[1].w, minmat[2].w) + minmat[3].w;
        judge();
        //5
        ans[anslen].h = max(minmat[0].h, minmat[1].h) + minmat[2].h + minmat[3].h;
        ans[anslen].w = max(minmat[0].w, minmat[1].w) + minmat[2].w + minmat[3].w;
        judge();
        //6
        if (minmat[0].w <= minmat[1].w && minmat[1].h <= minmat[3].h)
        {
            ans[anslen].h = max(minmat[0].h + minmat[1].h, minmat[2].h + minmat[3].h);
            ans[anslen].w = max(minmat[1].w + minmat[3].w, minmat[0].w + minmat[2].w);
            judge();
        }
    }

    for (int i = 0; i < 4; ++i)
        if (v[i])
        {
            v[i] = false;
            minmat[s].h = a[i].h;
            minmat[s].w = a[i].w;
            dfs(s+1);
            swap(minmat[s].h, minmat[s].w);
            dfs(s+1);
            v[i] = true;
        }
}

int main()
{
    freopen("packrec.in", "r", stdin);
    freopen("packrec.out", "w", stdout);
    for (int i = 0; i < 4; ++i) scanf("%d %d", &a[i].h, &a[i].w);
    memset(v, true, sizeof(v));
    anslen = 0;
    dfs(0);
    sort(ans, ans + anslen);
    printf("%d\n", ans[0].h * ans[0].w);
    printf("%d %d\n", ans[0].h, ans[0].w);
    for (int i = 1; i < anslen; ++i)
        if (ans[i].h != ans[i - 1].h || ans[i].w != ans[i - 1].w)
            printf("%d %d\n", ans[i].h, ans[i].w);
    return 0;
}