High priority queue of the server that receives the request from the server that receives the request.

High priority queue of the server that receives the request from the server that receives the request.

There is an NxN board. You need to put N cars on it. They cannot attack each other, and each car can only be placed in a specified rectangle.

 

Idea: first, cars cannot attack each other, so each row has only one vehicle in each column. We represent each vehicle in coordinates (x, y, the final requirement is that the x coordinates of any two vehicles are different, and the y coordinates of any two vehicles are different. Then, the x and y of each vehicle have their own range .....!!! X and y are independent of each other and will not affect each other !! So we only need to first find their own abscissa, and then find their own ordinate, right? At this time, the problem becomes quite simple. Our question becomes that there are n line segments on the X axis. You must take a point for each line segment, and the points for each line segment must be different from each other. Have we done this? Simply use the priority queue to get greedy.

 

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          #include using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)#define rrep(i,b,a) for(int i=(b);i>=(int)(a);--i)#define clr(a,x) memset(a,x,sizeof(a))#define mp make_pair#define eps 1e-10#define LL long long#define zero(x) (-eps < (x) && (x) < eps)const int maxn = 5000 + 5;int lx[maxn],rx[maxn],ly[maxn],ry[maxn];int X[maxn],Y[maxn];vector
          
            v[maxn];int n;bool Try(int *l, int * r,int * ret){ rep(i,0,n) v[i].clear(); rep(i,0,n) v[l[i]].push_back(i); priority_queue
           
             > q; rep(i,0,n){ rep(j, 0, v[i].size()){ int x = v[i][j]; q.push(mp(-r[x],x) ); } if(q.empty() || -q.top().first < i) return false; int c = q.top().second; q.pop(); ret[c]=i; } return true;}bool solve(){ if(!Try(lx,rx,X) || !Try(ly,ry,Y)) return false; rep(i,0,n) printf("%d %d\n",X[i]+1, Y[i]+1); return true;}int main(){ while(scanf("%d",&n),n){ rep(i,0,n){ scanf("%d%d%d%d",lx+i,ly+i,rx+i,ry+i); --lx[i];--ly[i];--rx[i];--ry[i]; } if(!solve()) puts("IMPOSSIBLE"); } return 0;}