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http://codeforces.com/gym/100463/attachments
Input
There are several test cases in each input file. The first line of each test case contains N (2 ≤ N ≤ 20), the number of points. The following N lines contain xi , yi , and ci (−1000 ≤ xi , yi , ≤ 1000, 0 ≤ ci ≤ 1) giving the x and y coordinates of the ith point. The ith point is red if ci = 0 and blue if ci = 1. The last line of input contains a zero.
Output
For each test case output the case number followed by the area of the smallest rectangle that satisfies the conditions above. If it is impossible output -1 instead. Follow the format in the sample output.
Sample Input
7 -10 0 0 -1 0 0 1 0 0 10 0 0 -1 -1 0 1 1 0 0 0 1 7 -4 0 0 -2 0 0 2 0 0 4 0 0 -3 0 1 0 0 1 3 0 1 0
Sample Output
Case 1: 9 Case 2: -1
题意
一个坐标系上面有n个点，要求找到一个矩形，使得能够框住一半的红点，不框进任何一个蓝点，求最小矩形面积
题解：
用dfs生成子集，并且判定矩形是否框住了蓝点。注意AC的代码中，是在所有至少有一半红点且无蓝点的矩形中寻找最小，所以最终如果有的话得到的一定是正好有一半红点的最小矩形。
//debug ：仍然不知道为什么第二种代码第一例就WA了
代码：
AC:
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <ctime> 5 #include <iostream> 6 #include <algorithm> 7 #include <set> 8 #include <vector> 9 #include <sstream> 10 #include <queue> 11 #include <typeinfo> 12 #include <fstream> 13 #include <map> 14 #include <stack> 15 typedef long long ll; 16 using namespace std; 17 //freopen("D.in","r",stdin); 18 //freopen("D.out","w",stdout); 19 #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) 20 #define test freopen("test.txt","r",stdin) 21 const int maxn=2501; 22 #define mod 1000000009 23 #define eps 1e-9 24 const int inf=0x3f3f3f3f; 25 const ll infll = 0x3f3f3f3f3f3f3f3fLL; 26 inline ll read() 27 { 28 ll x=0,f=1;char ch=getchar(); 29 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 30 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 31 return x*f; 32 } 33 //************************************************************************************** 34 35 int n; 36 int num1,num2; 37 struct node 38 { 39 int x,y; 40 }; 41 node a[40]; 42 node b[40]; 43 int ans; 44 bool ok(int x,int xx,int y,int yy) 45 { 46 for(int i=0;i<num2;i++) 47 if(b[i].x<=x&&b[i].x>=xx&&b[i].y<=y&&b[i].y>=yy) 48 return 0; 49 return 1; 50 } 51 52 void dfs(int t,int pre,int xmax,int xmin,int ymax,int ymin) 53 { 54 if(t>=num1/2) 55 { 56 if(ok(xmax,xmin,ymax,ymin)) 57 ans=min(ans,(xmax-xmin)*(ymax-ymin)); 58 return; 59 } 60 for(int i=pre+1;i<num1;i++) 61 dfs(t+1,i,max(a[i].x,xmax),min(xmin,a[i].x),max(ymax,a[i].y),min(ymin,a[i].y)); 62 } 63 int main() 64 { 65 int t=1; 66 while(cin>>n) 67 { 68 if(n==0) 69 break; 70 ans=inf; 71 num1=num2=0; 72 for(int i=0;i<n;i++) 73 { 74 int x=read(),y=read(),z=read(); 75 if(z==0) 76 a[num1].x=x,a[num1++].y=y; 77 else 78 b[num2].x=x,b[num2++].y=y; 79 } 80 dfs(0,-1,-inf,inf,-inf,inf); 81 if(ans!=inf) 82 printf("Case %d: %d\n",t++,ans); 83 else 84 printf("Case %d: -1\n",t++); 85 } 86 }
WA on first test:
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <ctime> 5 #include <iostream> 6 #include <algorithm> 7 #include <set> 8 #include <vector> 9 #include <sstream> 10 #include <queue> 11 #include <typeinfo> 12 #include <fstream> 13 #include <map> 14 #include <stack> 15 typedef long long ll; 16 using namespace std; 17 #define maxn 2000001 18 #define mod 10007 19 #define eps 1e-9 20 const int inf=0x3f3f3f3f; 21 const ll infll = 0x3f3f3f3f3f3f3f3fLL; 22 inline int read() 23 { 24 ll x=0,f=1;char ch=getchar(); 25 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 26 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 27 return x*f; 28 } 29 inline void out(int x) { 30 if(x>9) out(x/10); 31 putchar(x%10+‘0‘); 32 } 33 //************************************************************************************** 34 int n; 35 struct point 36 { 37 int x,y,is_blue; 38 }point[100]; 39 40 int init() 41 { 42 scanf("%d",&n); 43 for(int i=0;i<n;i++) 44 scanf("%d %d %d",&point[i].x,&point[i].y,&point[i].is_blue); 45 return n; 46 } 47 struct S 48 { 49 bool vis[100]; 50 int minS; 51 int minx,miny; 52 int maxx,maxy; 53 bool check() 54 { 55 int cnt_red=0,cnt_blue=0; 56 for(int i=0;i<n;i++) 57 { 58 int x,y,isblue; 59 x=point[i].x;y=point[i].y;isblue=point[i].is_blue; 60 if(x>=minx&&x<=maxx&&y>=miny&&y<=maxy) 61 { 62 if(isblue==1) cnt_blue++; 63 else cnt_red++; 64 } 65 } 66 if(cnt_blue==0&&cnt_red==n/2) return 1; 67 else return 0; 68 } 69 void dfs(int k,int d) 70 { 71 if(d==n/2) 72 { 73 minx=0x3f3f3f3f,miny=0x3f3f3f3f; 74 maxx=-1,maxy=-1; 75 for(int i=0;i<n;i++) 76 { 77 if(vis[i]) 78 { 79 int x=point[i].x,y=point[i].y; 80 minx=min(minx,x);miny=min(miny,y); 81 maxx=max(maxx,x);maxy=max(maxy,y); 82 } 83 } 84 int flag=check(); 85 if(flag==0) return ; 86 int dx=maxx-minx;int dy=maxy-miny; 87 if(minS==-1) minS=dx*dy; 88 else minS=min(minS,dx*dy); 89 return ; 90 } 91 if(k==n) return; 92 if(!point[k].is_blue) 93 { 94 vis[k]=1; 95 dfs(k+1,d+1); 96 } 97 vis[k]=0; 98 dfs(k+1,d); 99 } 100 void solve() 101 { 102 minS=-1; 103 memset(vis,0,sizeof(vis)); 104 dfs(0,0); 105 } 106 }S; 107 int main() 108 { 109 int Cases=1; 110 while(init()) 111 { 112 S.solve(); 113 if(S.minS!=-1) 114 printf("Case %d: %d\n",Cases,S.minS); 115 else 116 printf("Case %d: -1\n",Cases); 117 Cases++; 118 } 119 return 0; 120 }
Codeforces Gym 100463D Evil DFS
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原文地址：http://www.cnblogs.com/diang/p/4747089.html
