HDU 5601 N*M bulbs

Problem Description

N*M bulbs are in a rectangle, some are on, and some are off.

in order to save electricity, you should turn off all the lights, but you're lazy.
coincidentally,a passing bear children paper(bear children paper means the naughty boy), who want to pass here from the top left light bulb to the bottom right one and leave.

he starts from the top left light and just can get to the adjacent one in one step.
But after all,the bear children paper is just a bear children paper. after leaving a light bulb to the next one, he must touch the switch, which will change the status of the light.

your task is answer whether it's possible or not to finishing turning off all the lights, and make bear children paper also reach the bottom right light bulb and then leave at the same time.

 

Input

The first line of the input file contains an integer T, which indicates the number of test cases.

For each test case, there are n+1 lines.

The first line of each test case contains 2 integers n,m.

In the following n line contains a 01 square, 0 means off and 1 means on.

* T≤10
* N,M≤1000

 

Output

There should be exactly T lines in the output file.

The i-th line should only contain "YES" or "NO" to answer if it's possible to finish.

 

Sample Input

11 51 0 0 0 0

 

Sample Output

YES
Hint
Child's path is: (1,1)(1,2)(1,3)(1,2)(1,3)(1,4)(1,5)(4,5)all switches are touched twice except the first one.
 从左上角下去能改变的灯数和一开始亮着的灯数之间要满足这种关系才能到达
#include<map>#include<cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>using namespace std;typedef long long LL;const int maxn = 1e5 + 10;int T, n, m, x, sum;int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        sum = 0;        for (int i = 0; i < n;i++)            for (int j = 0; j < m; j++) { scanf("%d", &x); sum += x; }        if ((n + m + sum) % 2 == 1)            printf("YES\n"); else printf("NO\n");    }    return 0;}