Hdu4146

Flip Game

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1842    Accepted Submission(s): 607


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 (xi, yi) are given from input, which means that both pieces in row xi and pieces in column yi will be flipped (Note that piece (xi, yi) 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 (xi, yi), represent a command (1 <= N <= 1000, 0 <= Q <= 100000, 1 <= xi, yi <= 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

题意:给出一个有黑子和白子的图,有多次转化,输入X,Y就是把x行和y列的翻转。

求最后白子的数量。

  1. #include<stdio.h>  
  2. #include<string.h>  
  3. int main()  
  4. {  
  5.     int s,n,m,x,y,i,j,count,sum=1;  
  6.     char str[1001][1001];  
  7.     int a[1001],b[1001];  
  8.     scanf("%d",&s);  
  9.     while(s--)  
  10.     {  
  11.         memset(a,0,sizeof(a));  
  12.         memset(b,0,sizeof(b));  
  13.         count=0;  
  14.         scanf("%d",&n);  
  15.         for(i=0;i<n;i++)  
  16.         {  
  17.             scanf("%s",str[i]);//**这里需要解释下,二维数组输入str[i][j]时,首先读入的是str[0][j]**//  
  18.         }  
  19.         scanf("%d",&m);  
  20.         for(i=0;i<m;i++)  
  21.         {  
  22.             scanf("%d %d",&x,&y);//**代表第x行,第y列**//  
  23.             x--,y--;//**因为数组下表是从0开始的//**printf("%d %d\n",x,y);**//*//  
  24.             a[x]=1-a[x];//**x的横坐标改变**//  
  25.             b[y]=1-b[y];//**y的纵坐标改变**//  
  26.         }  
  27.         for(i=0;i<n;i++)  
  28.         {  
  29.             for(j=0;j<n;j++)  
  30.             {  
  31.                 if(a[i]+b[j]==1)//**判断改变次数的奇偶性<span style="font-size: 16px;">**//</span>  
  32.                 {  
  33.                     if(str[i][j]=='b') count++;  
  34.                 }  
  35.                 else  
  36.                 {  
  37.                     if(str[i][j]=='w') count++;  
  38.                 }  
  39.             }  
  40.         }  
  41.         printf("Case #%d: %d\n",sum++,count);  
  42.     }  
  43.     return 0;  
  44. }  
内容概要:该论文探讨了一种基于粒子群优化(PSO)的STAR-RIS辅助NOMA无线通信网络优化方法。STAR-RIS作为一种新型可重构智能表面,能同时反射和传输信号,与传统仅能反射的RIS不同。结合NOMA技术,STAR-RIS可以提升覆盖范围、用户容量和频谱效率。针对STAR-RIS元素众多导致获取完整信道状态信息(CSI)开销大的问题,作者提出一种在不依赖完整CSI的情况下,联合优化功率分配、基站波束成形以及STAR-RIS的传输和反射波束成形向量的方法,以最大化总可实现速率并确保每个用户的最低速率要求。仿真结果显示,该方案优于STAR-RIS辅助的OMA系统。 适合人群:具备一定无线通信理论基础、对智能反射面技术和非正交多址接入技术感兴趣的科研人员和工程师。 使用场景及目标:①适用于希望深入了解STAR-RIS与NOMA结合的研究者;②为解决无线通信中频谱资源紧张、提高系统性能提供新的思路和技术手段;③帮助理解PSO算法在无线通信优化问题中的应用。 其他说明:文中提供了详细的Python代码实现,涵盖系统参数设置、信道建模、速率计算、目标函数定义、约束条件设定、主优化函数设计及结果可视化等环节,便于读者理解和复现实验结果。此外,文章还对比了PSO与其他优化算法(如DDPG)的区别,强调了PSO在不需要显式CSI估计方面的优势。
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