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NENU ACM 2014-05-18 A题 && NENU 1081 Lucky

时间：2014-05-23 09:01:25 阅读：304 评论：0 收藏：0 [点我收藏+]

A - Lucky

Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu

Description

We say a n*m matrix r is lucky if and only if for every i, j, k, l(0 <= i < k < n, 0 <= j < l < m) r[i][j] + r[k][l] <= r[i][l] + r[k][j], and now you need to decide whether a matrix is luck.

Input

The input contains several test cases. The first line of each case contains two positive integers n and m(0 < n, m <= 2000), And then follow n lines and each line contains m integers. All of the integers are no more than 10^9 and no less than -10^9.

Output

If the matrix is lucky, please print “yes” without double quotation and if not print “no” without double quotation.

Sample Input

1 1
626
3 4
1382 1409 97 606
1548 1708 1973 1114
1626 381 75 799

Sample Output

yes
no

让人egg-pain的A题，我本来的思路是用两层循环找到一个点，再用两层循环找到另一个点，以这两个点为对角点建立一个矩形，依次判断条件 r[i][j] + r[k][l] <= r[i][l] + r[k][j] 是否成立，不过我很快发现这样的算法复杂度高达10^12（xiejj也用事实证明了暴搜行不通）。后来从AC的代码中发现了一条诡异的性质：如果一个大矩形的每个子矩形都满足上述条件，那么这个大矩形本身也满足这个条件，我自己也证明出来了，见页面底部。所以只需要用两层循环找到一个点，以这个点做单位矩阵，对每个单位矩阵判断就可以了。唉，不提了，说多了都是泪。。。。。AC过的代码如下：

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstdio>
 5 #include<math.h>
 6 #include<cstdlib>
 7 using namespace std;
 8 int r[2010][2010];
 9 bool flag=1;
10 int main() {
11     int n,m,i,j,k,l;
12     while(~scanf("%d%d",&n,&m)) {
13   if(n==0&&m==0) break;
14         for(i=0; i<n; i++)
15             for(j=0; j<m; j++)
16                 scanf("%d",&r[i][j]);
17         for(i=0,k=i+1;k<n&&i<n; i++)
18            for(j=0,l=j+1;flag==1&&l<m&&j<m; j++)
19                 //for(k=i+1; k<n; k++)
20                   //  for(l=j+1; flag==1&&l<m; l++)
21                         if(r[i][j] + r[k][l] > r[i][l] + r[k][j]) flag=0;
22 
23         if(flag==1) printf("yes\n");
24         else printf("no\n");
25     }
26 
27 
28     return 0;
29 }