山东省第一届ACM省赛 H SDUT 2158 Hello World!(穷举)
Hello World!
Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^
题目描述
“We need a programmer to help us for some projects. If you show us that you or 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 very simple.”
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 marked element, return a marked element whose row and column are larger than the showed element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still multiple solutions, return the element 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?
输入
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 row and 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.
输出
示例输入
3 1 2 2 3 2 30
示例输出
Case 1: 2 3 -1 -1 -1 -1
#include <bits/stdc++.h>
using namespace std;
struct list1{int x,y;}a[1111],b[1111];
bool cmp(list1 a,list1 b)
{if(a.x!=b.x) return a.x<b.x;return a.y<b.y;
}
int main()
{int n,k=1;while(scanf("%d",&n),n){for(int i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y),b[i].x=a[i].x,b[i].y=a[i].y;sort(a+1,a+1+n,cmp);printf("Case %d:\n",k++);for(int i=1;i<=n;i++){int f=0;for(int j=1;j<=n;j++)if(a[j].x>b[i].x && a[j].y>b[i].y){printf("%d %d\n",a[j].x,a[j].y);f=1;break;}if(!f) printf("-1 -1\n");}printf("\n");}return 0;
}
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