ZOJ Problem Set - 2480 Simplest Task in Windows

发布时间：2020-12-13 22:27:20 所属栏目：Windows 来源：网络整理

导读：Simplest Task in Windows Time Limit:? 2 Seconds ????? Memory Limit:? 65536 KB 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

Simplest Task in Windows

Time Limit:?2 Seconds ????? Memory Limit:?65536 KB

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.

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.

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.

Sample Input

1
0 0 5 5
3
4 1
5 1
6 1
0

Sample Output

0
0
-1

来源：?http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2480

方法一：我想的方法

 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",&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("%dn",j);
32                         flag=1;
33                         break;
34                     }
35                 }
36                 if(!flag) printf("-1n");
37             }
38             scanf("%d",&N);
39     }
40     return 0;
41 }

1、因为漏写了那个break纠结了蛮久的，所以以后碰到循环都稍微想一想什么时候该停下来

2、因为忘记初始化win数组也纠结了一会，所以以后碰到数组都要记得初始化

3、因为窗口输入顺序是back to front，注意逆序寻找窗口

方法二：

 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",&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("%dn",win[cx][cy]);
34                     flag=1;
35                 }
36                 if(!flag) printf("-1n");
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

（编辑：李大同）

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