Flip Game

Time Limit: 15000/5000 MS (Java/Others)Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1789Accepted Submission(s): 585

Problem Description Flip game is played on a square N*N field with two-sided pieces placed on each of its N^2 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. The rows are numbered with integers from 1 to N upside down; the columns are numbered with integers from 1 to N from the left to the right. Sequences of commands (x_i, y_i) are given from input, which means that both pieces in row x_i and pieces in column y_i will be flipped (Note that piece (x_i, y_i) will be flipped twice here). Can you tell me how many white pieces after sequences of commands?
Consider the following 4*4 field as an example:

bwww
wbww
wwbw
wwwb

Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up.
Two commands are given in order: (1, 1), (4, 4). Then we can get the final 4*4 field as follows:

bbbw
bbwb
bwbb
wbbb

So the answer is 4 as there are 4 white pieces in the final field.

Input The first line contains a positive integer T, indicating the number of test cases (1 <= T <= 20).
For each case, the first line contains a positive integer N, indicating the size of field; The following N lines contain N characters each which represent the initial field. The following line contain an integer Q, indicating the number of commands; each of the following Q lines contains two integer (x_i, y_i), represent a command (1 <= N <= 1000, 0 <= Q <= 100000, 1 <= x_i, y_i <= N).

Output For each case, please print the case number (beginning with 1) and the number of white pieces after sequences of commands.
Sample Input

2 4 bwww wbww wwbw wwwb 2 1 1 4 4 4 wwww wwww wwww wwww 1 1 1
Sample Output

Case #1: 4 Case #2: 10
Author Hzwu@SWJTU
Source 2011百校聯動“菜鳥杯”程序設計公開賽 題目大意：在一個x*y的方陣裡分為白和黑兩種顏色，題目說輸出1 1，4 4代表改變一次第一行第一列和第四行第四列。（如果Xi(代表X的橫坐標下表)與Yi（Y的縱坐標）相等時改變兩次）。最後要輸出的結果是這個方針中有多少白色塊。

 

解題思路：這個題目中時間要求特別嚴格，超時n多次~~~需要采用位運算進行次數的處理。（如果翻轉的次數為偶數次就不需要進行處理，因為翻轉了也要在翻轉回來）

還要就是輸入！！！剛開始采用的是字符輸入的超時！改成字符串輸入A掉了，祝你們好運~~~~

詳見代碼。

 

#include 
#include 
#include 

using namespace std;

char ch[1010][1010];
int vh[1010],vl[1010];
int n;

int main()
{
    int t;
    int q=1;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        memset(vh,0,sizeof(vh));
        memset(vl,0,sizeof(vl));
        for (int i=1; i<=n; i++)
        {
            scanf("%s",ch[i]+1);
            //getchar();
            //for (int j=1; j<=n; j++)
            //{
            //    ch[i][j]=getchar();
            //}
        }
        int k;
        scanf("%d",&k);
        while (k--)
        {
            int x,y;
            scanf("%d%d",&x,&y); 
            vh[x]^=1;
            vl[y]^=1;
        }
        int ans=0;
        for (int i=1; i<=n; i++)
        {
            for (int j=1; j<=n; j++)
            {
                if(vh[i]+vl[j]==1)
                {
                    if (ch[i][j]=='b')
                        ans++;
                }
                else if (ch[i][j]=='w')
                    ans++;
            }
        }
        //return ans;
        //turn();
        printf ("Case #%d: %d\n",q++,ans);
    }
    return 0;
}