USACO Packing Rectangles
1、#include <algorithm> 里面有自带的min()和max()函数
2、布局五实在是看不懂
3、使用了vector
vector<Rectangle> v;
v.push_back(rect);
sort(v.begin(), v.end(),cmp1);//按面积大小排序
vector<Rectangle>::iterator it = v.begin();
while( (it+1)!=v.end()){
if( (it->height * it->width) < ((it+1)->height * (it+1)->width)) break;//取最小面积
it++;
}
4、过滤重复的值：
vector<Rectangle>::iterator it2 = v.begin();
printf("%d %d\n", it2->width, it2->height);
for(it2++; it2 <= it; it2++){
if(it2->width == (it2-1)->width) continue;//相等的值不输出
printf("%d %d\n", it2->width, it2->height);
}
从http://blog.csdn.net/lucienduan/article/details/17022119
的代码：
/*
ID: nenusb1
PROG: packrec
LANG: C++
*/
/*
*深度优先遍历，找出最优解
**/
#include<iostream>
#include<fstream>
#include<vector>
#include<algorithm>
using namespace std;
typedef struct Rectangle
{
int width;
int height;
}Rectangle;
bool cmp1(Rectangle rect1,Rectangle rect2)
{
return rect1.height*rect1.width<rect2.height*rect2.width;
}
bool cmp2(Rectangle rect1,Rectangle rect2)
{
return rect1.width<rect2.width;
}
int main()
{
freopen("packrec.in","r",stdin);
freopen("packrec.out","w",stdout);
Rectangle rects[4];
int width,height;
for (int i=0;i<4;i++)
{
scanf("%d %d",&width,&height);
rects[i].width = min(width,height);
rects[i].height = max(width,height);
}
//测试6种基本布局（实际只有5中，4,5是相同的）
//对每一种布局依次遍历每种矩阵的摆放方式
// 每个矩阵都可能旋转90度，共4！*2^4=384种
// 找出最小的大矩阵
vector<Rectangle> v;
Rectangle rect;
for (int i=0;i<4;i++)
{
int icount=1;
while(icount<=2){
if(icount==2)//矩阵旋转90度
{
int temp=rects[i].height;
rects[i].height=rects[i].width;
rects[i].width=temp;
}
for (int j=0;j<4;j++)
{
if(i==j)
continue;
int jcount=1;
while(jcount<=2){
if(jcount==2)
{
int temp=rects[j].height;
rects[j].height=rects[j].width;
rects[j].width=temp;
}
for (int k=0;k<4;k++)
{
if(i==k || j==k)
continue;
int kcount=1;
while(kcount<=2){
if(kcount==2)
{
int temp=rects[k].height;
rects[k].height=rects[k].width;
rects[k].width=temp;
}
for (int t=0;t<4;t++)
{
if(i==t || j==t || k==t)
continue;
int tcount=1;
while(tcount<=2){
if(tcount==2)
{
int temp=rects[t].height;
rects[t].height=rects[t].width;
rects[t].width=temp;
}
//布局一
height=max(max(max(rects[i].height,rects[j].height),rects[k].height),rects[t].height);
width=rects[i].width+rects[j].width+rects[k].width+rects[t].width;
rect.height=max(height,width);
rect.width=min(height,width);
v.push_back(rect);
//布局二
if(rects[t].height>rects[j].width+rects[k].width)
{
height=max(max(rects[i].height,rects[j].height),rects[k].height)+rects[t].width;
width=max(rects[i].width+rects[j].width+rects[k].width,rects[t].height);
rect.height=max(height,width);
rect.width=min(height,width);
v.push_back(rect);
}
//布局三
if(rects[t].height>rects[j].width)
{
height=max(max(rects[i].height,rects[j].height)+rects[t].width,rects[k].height);
width=max(rects[i].width+rects[j].width,rects[t].height)+rects[k].width;
rect.height=max(height,width);
rect.width=min(height,width);
v.push_back(rect);
}
//布局四
height=max(max(rects[i].height,rects[k].height),rects[j].height+rects[t].height);
width=max(rects[j].width,rects[t].width)+rects[i].width+rects[k].width;
rect.height=max(height,width);
rect.width=min(height,width);
v.push_back(rect);
//布局五，最麻烦的,左右两边其实是对称的情况，所以只考虑一边就行
if(rects[k].height>=rects[t].height)
{
if(rects[i].height+rects[t].height>rects[k].height && rects[i].width<=rects[t].width)
{//这个条件通过镜像反射代表了两个条件
width=max(rects[i].width+rects[j].height,rects[k].width+rects[t].width);
}
}else if(rects[j].width+rects[k].height<=rects[t].height){
width=max(max(rects[j].height,rects[k].width)+rects[t].width,rects[i].width);
}else{
width=max(rects[i].width,rects[t].width)+max(rects[j].height,rects[k].width);
}
height=max(rects[i].height+rects[t].height,rects[j].width+rects[k].height);
rect.height=max(height,width);
rect.width=min(height,width);
v.push_back(rect);
tcount++;
}
}
kcount++;
}
}
jcount++;
}
}
icount++;
}
}
sort(v.begin(), v.end(),cmp1);//按面积大小排序
vector<Rectangle>::iterator it = v.begin();
// while(it< v.end()){
// printf("%d %d\n", it->width, it->height);
// it++;
// }
// printf("-------------------");
it = v.begin();
while( (it+1)!=v.end()){
if( (it->height * it->width) < ((it+1)->height * (it+1)->width)) break;//取最小面积
it++;
}
sort(v.begin(), it+1, cmp2);//将最小面积的方案按宽度从小到大排序
printf("%d\n",it->height * it->width);//最小面积的值
vector<Rectangle>::iterator it2 = v.begin();
printf("%d %d\n", it2->width, it2->height);
for(it2++; it2 <= it; it2++){
if(it2->width == (it2-1)->width) continue;//相等的值不输出
printf("%d %d\n", it2->width, it2->height);
}
return 0;
}