标签：lines without using basic and div esc 6.2 std
Four rectangles are given. Find the smallest enclosing (new) rectangle into which these four may be fitted without overlapping. By smallest rectangle, we mean the one with the smallest area.
All four rectangles should have their sides parallel to the corresponding sides of the enclosing rectangle. Figure 1 shows six ways to fit four rectangles together. These six are the only possible basic layouts, since any other layout can be obtained from a basic layout by rotation or reflection. Rectangles may be rotated 90 degrees during packing.
There may exist several different enclosing rectangles fulfilling the requirements, all with the same area. You must produce all such enclosing rectangles.
Four lines, each containing two positive space-separated integers that represent the lengths of a rectangle‘s two sides. Each side of a rectangle is at least 1 and at most 50.
1 2 2 3 3 4 4 5
The output file contains one line more than the number of solutions. The first line contains a single integer: the minimum area of the enclosing rectangles. Each of the following lines contains one solution described by two numbers p and q with p<=q. These lines must be sorted in ascending order of p, and must all be different.
40 4 10 5 8
————————————————————————————————————————————————题解
我们枚举每一种情况
1.将编号为1 2 3 4的全排列
2.将排列后的1情况的每一个矩形枚举转还是不转
将1、2做完之后手动模拟6种情况即可
1 /* 2 ID: ivorysi 3 LANG: C++ 4 PROG: packrec 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <queue> 10 #include <set> 11 #include <vector> 12 #include <algorithm> 13 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 14 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 15 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 16 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 17 #define inf 0x5f5f5f5f 18 #define ivorysi 19 #define mo 97797977 20 #define hash 974711 21 #define base 47 22 #define fi first 23 #define se second 24 #define pii pair<int,int> 25 #define esp 1e-8 26 typedef long long ll; 27 using namespace std; 28 int n,area=inf; 29 vector<pii> v; 30 void record(pii t) { 31 if(t.fi>t.se) swap(t.fi,t.se); 32 if(t.fi*t.se==area) v.push_back(t); 33 else if(t.fi*t.se<area) { 34 v.clear(); 35 area=t.fi*t.se; 36 v.push_back(t); 37 } 38 } 39 struct data { 40 int l,r; 41 }squ[5],rec[5]; 42 inline void rotate(data &a) { 43 swap(a.l,a.r); 44 } 45 bool used[5]; 46 void calc() { 47 pii w; 48 //fi是竖边,se是横边 49 //case 1 50 w.fi=0; 51 siji(i,1,4) w.fi=max(w.fi,rec[i].l); 52 siji(i,1,4) w.se+=rec[i].r; 53 record(w); 54 //case 2 55 int temp2=0; 56 siji(i,1,3) temp2=max(rec[i].l,temp2); 57 w.fi=rec[4].l+temp2; 58 w.se=0; 59 siji(i,1,3) w.se+=rec[i].r; 60 w.se=max(w.se,rec[4].r); 61 record(w); 62 //case 3 63 w.fi=max(rec[1].l,rec[2].l)+rec[3].l; 64 w.fi=max(w.fi,rec[4].l); 65 w.se=max(rec[3].r,rec[1].r+rec[2].r)+rec[4].r; 66 record(w); 67 //case 4,5 68 w.fi=max(rec[1].l,rec[2].l); 69 w.fi=max(w.fi,rec[3].l+rec[4].l); 70 w.se=rec[1].r+rec[2].r; 71 w.se+=max(rec[3].r,rec[4].r); 72 record(w); 73 //case 6 74 // 1 2 75 // 3 4 76 w.fi=max(rec[1].l+rec[3].l,rec[2].l+rec[4].l); 77 w.se=rec[3].r+rec[4].r; 78 // 1与2 79 if(rec[1].l+rec[3].l>rec[4].l) w.se=max(w.se,rec[1].r+rec[2].r); 80 // 2与3 81 if(rec[3].l>rec[4].l) w.se=max(w.se,rec[2].r+rec[3].r); 82 // 1与4 83 if(rec[4].l>rec[3].l) w.se=max(w.se,rec[1].r+rec[4].r); 84 // 1 或 2 特别长 85 w.se=max(w.se,rec[1].r); 86 w.se=max(w.se,rec[2].r); 87 record(w); 88 } 89 void dfs1(int k) { 90 if(k>4) { 91 calc(); 92 return; 93 } 94 dfs1(k+1);//不转这个 95 rotate(rec[k]); 96 dfs1(k+1);//转这个 97 rotate(rec[k]); 98 } 99 void dfs(int k) { 100 if(k>4) { 101 dfs1(1); 102 } 103 siji(i,1,4) { 104 if(!used[i]) { 105 rec[k]=squ[i]; 106 used[i]=1; 107 dfs(k+1); 108 used[i]=0; 109 } 110 } 111 } 112 void solve() { 113 siji(i,1,4) scanf("%d%d",&squ[i].l,&squ[i].r); 114 dfs(1); 115 sort(v.begin(),v.end()); 116 vector<pii>::iterator it=unique(v.begin(),v.end()); 117 v.erase(it,v.end()); 118 printf("%d\n",area); 119 siji(i,0,v.size()-1) { 120 printf("%d %d\n",v[i].fi,v[i].se); 121 } 122 } 123 int main(int argc, char const *argv[]) 124 { 125 #ifdef ivorysi 126 freopen("packrec.in","r",stdin); 127 freopen("packrec.out","w",stdout); 128 #else 129 freopen("f1.in","r",stdin); 130 #endif 131 solve(); 132 return 0; 133 }
标签：lines without using basic and div esc 6.2 std
原文地址：http://www.cnblogs.com/ivorysi/p/6662962.html
