HDU4146 Flip Game

Flip Game

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 824    Accepted Submission(s): 283

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

/*
//代码一：-----TLE
#include<stdio.h>
#include<string.h>
char str[1005][1005];
int cnt[1005][1005];

int main()
{
    int T,i,j,k,n,m,sum,row,col;
    scanf("%d",&T);
    for(k=1;k<=T;++k)
    {
        scanf("%d",&n);
        getchar();
        for(i=0;i<n;++i)
        {
            for(j=0;j<n;++j)
            {
                scanf("%c",&str[i][j]);
                cnt[i][j]=0;
            }
            getchar();
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&row,&col);
            --row;
            --col;
            for(i=0;i<n;++i)
            {
                ++cnt[row][i];
                ++cnt[i][col];
            }
        }
        sum=0;
        for(i=0;i<n;++i)
            for(j=0;j<n;++j)
            {
                if(str[i][j]=='b'&&(cnt[i][j]&1))
                    ++sum;
                if(str[i][j]=='w'&&!(cnt[i][j]&1))
                    ++sum;
            }
        printf("Case #%d: %d\n",k,sum);
    }
    return 0;
}
*/

//代码二：------WA----看半天看不出来哪错了，跪求路过的拯救
#include<stdio.h>
#include<string.h>
char str[1005][1005];

int main()
{
    int T,i,j,k,n,m,sum,row,col;
    scanf("%d",&T);
    for(k=1;k<=T;++k)
    {
        sum=0;
        scanf("%d",&n);
        getchar();
        for(i=0;i<n;++i)
        {
            for(j=0;j<n;++j)
            {
                scanf("%c",&str[i][j]);
                if(str[i][j]=='w')
                    ++sum;
            }
            getchar();
        }
        scanf("%d",&m);
        while(m--)   //输入一条指令更新计算一次
        {
            scanf("%d%d",&col,&row);
            --row;
            --col;
            for(i=0;i<n;++i)
            {
                if(str[row][i]=='b')
                {
                    ++sum;
                    str[row][i]='w';
                }
                else
                {
                    --sum;
                    str[row][i]='b';
                }
                if(str[i][col]=='b')
                {
                    ++sum;
                    str[i][col]='w';
                }
                else
                {
                    --sum;
                    str[i][col]='b';
                }
            }
        }
        printf("Case #%d: %d\n",k,sum);
    }
    return 0;
}

/*
//代码三：-----AC
#include<cstdio>
#include<cstring>
char grid[1005][1005];
int col[1005],row[1005];  //分别记录该列和该行变化的次数为奇数次还是偶数次，用1和0表示(异或运算)
int main()
{
    int T,r,n,Q,a,b,i,j;
    scanf("%d",&T);
    for(r=1;r<=T;++r)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%s",grid[i]);
        scanf("%d",&Q);
        memset(col,0,sizeof(col));
        memset(row,0,sizeof(row));
        for(i=0;i<Q;i++)
        {
            scanf("%d%d",&a,&b);
            --a;
            --b;
            row[a]^=1;
            col[b]^=1;
        }
        int res=0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if((row[i]+col[j])&1) 
                {
                    if(grid[i][j]=='b') 
                        res++;
                }
                else 
                {
                    if(grid[i][j]=='w') 
                        res++;
                }
            }
        }
        printf("Case #%d: %d\n",r,res);
    }
    return 0;
}
*/