Codeforces Round #340 (Div. 2)——D
2016-01-24 13:37
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题意:
给你3个点的坐标,然后问你最少用几条折线就可以把这3个点给连接起来,并且同时要满足这几条折线都平行于坐标轴。(具体详见样例)
思路:
直接模拟,要小心。。。
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
typedef __int64 ll;
typedef unsigned __int64 ULL;
#define inf 99999999
#define maxn 5
struct node{
ll x,y;
}a[maxn];
bool cmp(node a,node b){
if(a.x!=b.x) return a.x<b.x;
else return a.y<b.y;
}
int main(){
for(int i=0;i<3;i++){
scanf("%I64d%I64d",&a[i].x,&a[i].y);
}
sort(a,a+3,cmp);
if((a[0].x==a[1].x&&a[1].x==a[2].x)||(a[0].y==a[1].y&&a[1].y==a[2].y)){
printf("1\n");
}
//当有2个点在同一直线上的情况要小心,因为在范围之外的也是可行的!!
else if((a[0].x==a[1].x&&(a[2].y>=a[1].y||a[2].y<=a[0].y))||
(a[0].y==a[1].y&&(a[2].x<=a[0].x||a[2].x>=a[1].x))||
(a[0].y==a[2].y&&(a[1].x<=a[0].x||a[1].x>=a[2].x))||
(a[1].x==a[2].x&&(a[0].y<=a[1].y||a[0].y>=a[2].y))||
(a[1].y==a[2].y&&(a[0].x<=a[1].x||a[0].x>=a[2].x))){
printf("2\n");
}
else{
printf("3\n");
}
}
给你3个点的坐标,然后问你最少用几条折线就可以把这3个点给连接起来,并且同时要满足这几条折线都平行于坐标轴。(具体详见样例)
思路:
直接模拟,要小心。。。
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
typedef __int64 ll;
typedef unsigned __int64 ULL;
#define inf 99999999
#define maxn 5
struct node{
ll x,y;
}a[maxn];
bool cmp(node a,node b){
if(a.x!=b.x) return a.x<b.x;
else return a.y<b.y;
}
int main(){
for(int i=0;i<3;i++){
scanf("%I64d%I64d",&a[i].x,&a[i].y);
}
sort(a,a+3,cmp);
if((a[0].x==a[1].x&&a[1].x==a[2].x)||(a[0].y==a[1].y&&a[1].y==a[2].y)){
printf("1\n");
}
//当有2个点在同一直线上的情况要小心,因为在范围之外的也是可行的!!
else if((a[0].x==a[1].x&&(a[2].y>=a[1].y||a[2].y<=a[0].y))||
(a[0].y==a[1].y&&(a[2].x<=a[0].x||a[2].x>=a[1].x))||
(a[0].y==a[2].y&&(a[1].x<=a[0].x||a[1].x>=a[2].x))||
(a[1].x==a[2].x&&(a[0].y<=a[1].y||a[0].y>=a[2].y))||
(a[1].y==a[2].y&&(a[0].x<=a[1].x||a[0].x>=a[2].x))){
printf("2\n");
}
else{
printf("3\n");
}
}
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