CF判斷正方形和矩形
//本題總結:判斷特殊的的多邊形的時候..要抓住很特別的特徵..一定要能夠確定是該多邊形的條件.
//比如本題..出現了一個很特殊的直角梯形的情況..有了四個最小邊相等..然後存在一個直角的情況..
//但是..開始寫的時候的條件改成去判斷全部形成的角是不是存在四個直角的話..就不會出現這樣的情況了..
//有的時候..對角線會有比邊小的情形出現..
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Node
{
int x,y;
Node(int x1,int y1):x(x1),y(y1){}
Node(){};
};
int distan(Node a,Node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
bool judge(Node a, Node b ,Node c)
{
if( (a.x-b.x)*(c.x-b.x) == -(a.y-b.y)*(c.y-b.y))//判斷以角abc是否是直角
return 1;
return 0;
}
bool judge1(Node a, Node b ,Node c,Node d)//判斷是否是正方形.
{
int dis[10];
dis[1] = distan(a,b);
dis[2] = distan(a,c);
dis[3] = distan(a,d);
dis[4] = distan(b,c);
dis[5] = distan(b,d);
dis[6] = distan(c,d);
sort(dis+1,dis+7);
if(dis[1] == dis[2] && dis[2] == dis[3] && dis[3] == dis[4])//尋找最小的四條邊是否是相等.如果是正方形的話..
{ //必然能夠存在四條最小的邊就是邊長.
int m = judge(a,b,c) + judge(a,b,d) + judge(c,b,d) + judge(b,a,c) + judge(b,a,d) + judge(c,a,d)+ judge(a,c,b)+ judge(d,c,b)+ judge(d,c,a) + judge(a,d,b)+judge(a,d,c)+judge(b,d,c);//再次判斷是否存在四個直角
if(m == 4)return 1;
}
else return 0;
}
bool judge2(Node a, Node b ,Node c,Node d)
{
int dis[10];
dis[1] = distan(a,b);
dis[2] = distan(a,c);
dis[3] = distan(a,d);
dis[4] = distan(b,c);
dis[5] = distan(b,d);
dis[6] = distan(c,d);
sort(dis+1,dis+7);
if(dis[1] == dis[2] && dis[3] == dis[4])
{
int m = judge(a,b,c) + judge(a,b,d) + judge(c,b,d) + judge(b,a,c) + judge(b,a,d) + judge(c,a,d)+ judge(a,c,b)+ judge(d,c,b)+ judge(d,c,a) + judge(a,d,b)+judge(a,d,c)+judge(b,d,c);
if( m == 4)
return 1;
}
return 0;
}
int main()
{
Node node[10];
while(scanf("%d%d",&node[1].x,&node[1].y) != EOF)
{
for(int i = 2 ; i <= 8 ; i++)
{
scanf("%d%d",&node[i].x,&node[i].y);
}
bool vis[11];
bool wori = 0;
memset(vis,0,sizeof(vis));
int ans[111];
memset(ans,0,sizeof(ans));
for(int i = 1 ; i <= 8 ; i++)
{
if(vis[i])continue;
if(wori)break;
for(int j = 1 ; j <= 8 ; j++)
{
if(j == i || vis[j])continue;
if(wori)break;
for(int k = 1 ; k <= 8 ; k++)
{
if(k == i || k == j ||vis[k])continue;
if(wori)break;
for(int p = 1 ; p <= 8 ; p++)
{
if(p == i || p == j || p == k || vis[p])continue;
if(wori)break;
if(judge1(node[i],node[j],node[k],node[p]))
{
int t = 1;
for(int pp = 1 ; pp <= 8 ; pp++)
{
if(pp != i && pp != j && pp != k && pp != p)
{
ans[t++] = pp;
//printf("%d \n",pp);
}
}
if(judge2(node[ans[1]],node[ans[2]],node[ans[3]],node[ans[4]]))
{
vis[i] = 1;
vis[j] = 1;
vis[k] = 1;
vis[p] = 1;
printf("YES\n");
printf("%d %d %d %d\n",i,j,k,p);
printf("%d %d %d %d\n",ans[1],ans[2],ans[3],ans[4]);
wori = 1;
}
}
}
}
}
}
if(!wori)printf("NO\n");
}
}
//比如本題..出現了一個很特殊的直角梯形的情況..有了四個最小邊相等..然後存在一個直角的情況..
//但是..開始寫的時候的條件改成去判斷全部形成的角是不是存在四個直角的話..就不會出現這樣的情況了..
//有的時候..對角線會有比邊小的情形出現..
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Node
{
int x,y;
Node(int x1,int y1):x(x1),y(y1){}
Node(){};
};
int distan(Node a,Node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
bool judge(Node a, Node b ,Node c)
{
if( (a.x-b.x)*(c.x-b.x) == -(a.y-b.y)*(c.y-b.y))//判斷以角abc是否是直角
return 1;
return 0;
}
bool judge1(Node a, Node b ,Node c,Node d)//判斷是否是正方形.
{
int dis[10];
dis[1] = distan(a,b);
dis[2] = distan(a,c);
dis[3] = distan(a,d);
dis[4] = distan(b,c);
dis[5] = distan(b,d);
dis[6] = distan(c,d);
sort(dis+1,dis+7);
if(dis[1] == dis[2] && dis[2] == dis[3] && dis[3] == dis[4])//尋找最小的四條邊是否是相等.如果是正方形的話..
{ //必然能夠存在四條最小的邊就是邊長.
int m = judge(a,b,c) + judge(a,b,d) + judge(c,b,d) + judge(b,a,c) + judge(b,a,d) + judge(c,a,d)+ judge(a,c,b)+ judge(d,c,b)+ judge(d,c,a) + judge(a,d,b)+judge(a,d,c)+judge(b,d,c);//再次判斷是否存在四個直角
if(m == 4)return 1;
}
else return 0;
}
bool judge2(Node a, Node b ,Node c,Node d)
{
int dis[10];
dis[1] = distan(a,b);
dis[2] = distan(a,c);
dis[3] = distan(a,d);
dis[4] = distan(b,c);
dis[5] = distan(b,d);
dis[6] = distan(c,d);
sort(dis+1,dis+7);
if(dis[1] == dis[2] && dis[3] == dis[4])
{
int m = judge(a,b,c) + judge(a,b,d) + judge(c,b,d) + judge(b,a,c) + judge(b,a,d) + judge(c,a,d)+ judge(a,c,b)+ judge(d,c,b)+ judge(d,c,a) + judge(a,d,b)+judge(a,d,c)+judge(b,d,c);
if( m == 4)
return 1;
}
return 0;
}
int main()
{
Node node[10];
while(scanf("%d%d",&node[1].x,&node[1].y) != EOF)
{
for(int i = 2 ; i <= 8 ; i++)
{
scanf("%d%d",&node[i].x,&node[i].y);
}
bool vis[11];
bool wori = 0;
memset(vis,0,sizeof(vis));
int ans[111];
memset(ans,0,sizeof(ans));
for(int i = 1 ; i <= 8 ; i++)
{
if(vis[i])continue;
if(wori)break;
for(int j = 1 ; j <= 8 ; j++)
{
if(j == i || vis[j])continue;
if(wori)break;
for(int k = 1 ; k <= 8 ; k++)
{
if(k == i || k == j ||vis[k])continue;
if(wori)break;
for(int p = 1 ; p <= 8 ; p++)
{
if(p == i || p == j || p == k || vis[p])continue;
if(wori)break;
if(judge1(node[i],node[j],node[k],node[p]))
{
int t = 1;
for(int pp = 1 ; pp <= 8 ; pp++)
{
if(pp != i && pp != j && pp != k && pp != p)
{
ans[t++] = pp;
//printf("%d \n",pp);
}
}
if(judge2(node[ans[1]],node[ans[2]],node[ans[3]],node[ans[4]]))
{
vis[i] = 1;
vis[j] = 1;
vis[k] = 1;
vis[p] = 1;
printf("YES\n");
printf("%d %d %d %d\n",i,j,k,p);
printf("%d %d %d %d\n",ans[1],ans[2],ans[3],ans[4]);
wori = 1;
}
}
}
}
}
}
if(!wori)printf("NO\n");
}
}
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