題目描述
We know thatIvan gives Saya three problems to solve (Problem F), and this is the firstproblem.
“We need a programmer to help us for some projects. If you show us that youor one of your friends is able to program, you can pass the first hurdle.
I will give you a problem to solve. Since this is the first hurdle, it is verysimple.”
We all know that the simplest program is the “Hello World!”program. This is a problem just as simple as the
“Hello World!”
In a large matrix, there are some elements has been marked. For every markedelement, return a marked element whose row and column are larger than theshowed element’s row andcolumn respectively.
If there are multiple solutions, return the element whoserow is the smallest; and if there are still multiple solutions, return theelement whose column is the smallest. If there is no solution, return -1 -1.
Saya is not a programmer, so she comes to you for help
Can you solve this problem for her?
輸入
The inputconsists of several test cases.
The first line of input in each test case contains one integer N (0<N≤1000), which represents the number of marked element.
Each of the next N lines containing two integers r and c,represent the element’s rowand column. You can assume that 0<r,c≤300.
A marked element can be repeatedly showed.
The last case is followed by a line containing one zero.
輸出
For each case,print the case number (1, 2 …), andfor each element’s rowand column, output the result. Your output format should imitate the sampleoutput. Print a blank line after each test case.
示例輸入
3
1 2
2 3
2 3
0
示例輸出
Case 1:
2 3
-1 -1
-1 -1
/************************************
在一個矩陣中有一些標記元素（行 r 和列 c ），找出比這些元素行，列下標大的標記元素的行和列並輸出，如果有多個輸出行最小的，行相同輸出列最小的
簡單的數學題，排序，因爲每個元素有兩個下標（r，c），以行（r）爲主排序，若行（r）相同，則比較列（c）..然後判斷一下就行。
**************************************/
Code:
#include <stdio.h>
#include<algorithm>
#include <string.h>
using namespace std;
struct Point // 定義一個 元素 (標記) 的結構體。
{
int x;
int y;
}point[1005],p[1005];
bool cmp(Point a,Point b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{
int n,count_case = 1,i,j,k;
while(scanf("%d",&n)&&n)
{
for(i = 0;i<n;i++)
{
scanf("%d%d",&point[i].x,&point[i].y);
p[i].x = point[i].x;p[i].y = point[i].y;// 拷貝一份數據
}
std::sort(p,p+n,cmp); // 將拷貝數據備份
printf("Case %d:\n",count_case++);
for(i = 0;i<n;i++)
{
j = 0;
while((point[i].x>=p[j].x||point[i].y>=p[j].y)&&j<n)// 遍歷，如果標記元素 的 x y 比要顯示的小 則繼續，否則跳出循環
j++;
if(j<n)
printf("%d %d\n",p[j].x,p[j].y);// 找到了比 標記元素 的 x y 大的元素，輸出
else
printf("-1 -1\n");// 沒找到，輸出 -1 -1
}
printf("\n");
}
return 0;
}