建立两个数组同时存储原学校与目标学校。分别按优先原学校和优先目标学校排序,一一配对,验查是否可行。
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 5 struct node { 6 int x,y; 7 }a[500010],b[500010]; 8 9 bool cmp1 (node x,node y){ 10 if ((x.x<y.x)||(x.x==y.x&&x.y<y.y)) 11 return true ; 12 return false ; 13 } 14 bool cmp2 (node x,node y){ 15 if ((x.y<y.y)||(x.y==y.y&&x.x<y.x)) 16 return true ; 17 return false ; 18 } 19 20 int main (){ 21 int n; 22 while (cin>>n&&n){ 23 for (int i=0;i<n;i++){ 24 cin>>a[i].x>>a[i].y; 25 b[i].x=a[i].x; 26 b[i].y=a[i].y; 27 } 28 sort (a,a+n,cmp1); 29 sort (b,b+n,cmp2); 30 int flag=1; 31 for (int i=0;i<n;i++){//cout<<i<<":"<<a[i].x<<" "<<a[i].y<<" "<<b[i].x<<" "<<b[i].y<<endl; 32 if (a[i].x==b[i].y&&a[i].y==b[i].x) 33 continue ; 34 flag=0;break ; 35 } 36 if (flag) 37 cout<<"YES"<<endl; 38 else cout<<"NO"<<endl; 39 } 40 return 0; 41 }
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原文地址:http://www.cnblogs.com/gfc-g/p/3852988.html