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A typical windows platform may have several windows on the desktop. A user‘s operation may be as simple as a single click of the mouse. In the implementation of such a windows platform,one of the simplest tasks would be deciding which window had been clicked.
Given a serial of windows,for each mouse click,decide which window had been clicked. Please notice that windows may overlap,and the window on top would receive the click rather than others. And,we consider a window to be clicked even if the mouse is just on its edge/corner. For the sake of simplicity,we assume that a window will not be activated to the top by any click.
@H_944_35@Input
The first part of input is a serial of windows. First an Integer N (1 <= N <= 10) is given,inDicaTing the number of windows. Then N lines follow,each containing four Integers,x1,y1,x2,y2,(x1 < x2,y1 < y2) the coordinates of the upper-left and lower-right corners of a window. The windows are given in BACk-to-front order. N=0 signals the end of input.
The second part of input is a serial of mouse clicks. First an Integer M (1 <= M <= 50) is given,inDicaTing the number of mouse clicks. Then M lines follow,each containing two Integers,x and y,the coordinates of a mouse click.
@H_944_35@Output
For each mouse click in the input,output a single line containing the index (starTing from 0) of the window which receives the click. If there is no such window,output "-1" in a line instead.
@H_944_35@Sample Input
1
0 0 5 5
3
4 1
5 1
6 1
0
@H_944_35@Sample Output
0
0
-1
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int win[15][4]; 5 int main() 6 { int N; 7 scanf("%d",&N); 8 while(N>0){ 9 for(int i=0;i<N;i++) 10 for(int j=0;j<4;j++) 11 win[i][j]=-1; 12 for(int i=0;i<N;i++){ 13 for(int j=0;j<4;j++){ 14 int temp; 15 scanf("%d",&temp); 16 win[i][j]=temp; 17 } 18 } 19 int M; 20 bool flag=0; 21 scanf("%d",&@H_670_93@m); 22 for(int i=0;i<M;i++){ 23 flag=0; 24 int cx,cy; 25 scanf("%d",&cX); 26 scanf("%d",&cy); 27 for(int j=N-1;j>=0;j--){ 28 if(win[i][j]==-1) 29 conTinue; 30 if(cx<=win[j][2]&&cx>=win[j][0]&&cy<=win[j][3]&&cy>=win[j][1]){ 31 printf("%d\n",j); 32 flag=1; 33 break; 34 } 35 } 36 if(!flag) printf("-1\n"); 37 } 38 scanf("%d",&N); 39 } 40 return 0; 41 }
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int win[107][107]; 5 int main() 6 { int N; 7 scanf("%d",&N); 8 while(N>0){ 9 for(int i=0;i<107;i++){ 10 for(int j=0;j<107;j++) 11 win[i][j]=-1; 12 } 13 for(int i=0;i<N;i++){ 14 int l,d,r,u; 15 scanf("%d",&l); 16 scanf("%d",&d); 17 scanf("%d",&r); 18 scanf("%d",&u); 19 for(int j=l;j<=r;j++){ 20 for(int k=d;k<=u;k++) 21 win[j][k]=i; 22 } 23 } 24 int M; 25 bool flag=0; 26 scanf("%d",&@H_670_93@m); 27 for(int i=0;i<M;i++){ 28 flag=0; 29 int cx,cy; 30 scanf("%d",&cX); 31 scanf("%d",&cy); 32 if(win[cx][cy]!=-1){ 33 printf("%d\n",win[cx][cy]); 34 flag=1; 35 } 36 if(!flag) printf("-1\n"); 37 } 38 scanf("%d",&N); 39 } 40 return 0; 41 }
根据别人的思路打的，这个思路也不错，用二维数组模拟一块屏幕，但是因为题目没有提及coordinate的xy取值范围，所以win的大小其实不好拿捏，但是这个107能AC
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