God Save the i-th Queen

最新推荐文章于 2023-08-14 19:08:15 发布

Tian_Fei

最新推荐文章于 2023-08-14 19:08:15 发布

阅读量358

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/lois_123/article/details/45030703

本文探讨了在给定尺寸的棋盘上，已有若干皇后放置的情况下，如何计算还能放置皇后的空位数量。通过标记已占据的位置及受攻击威胁的行列和对角线，采用双层循环遍历的方法高效解决问题。

God Save the i-th Queen

Time Limit: 5000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

Submit Status PID: 4299

Did you know that during the ACM-ICPC World Finals a big chessboard is installed every year and is available for the participants to play against each other? In this problem, we will test your basic chess-playing abilities to verify that you would not make a fool of yourself if you advance to the World Finals.

During the yesterday’s Practice Session, you tried to solve the problem of N independent rooks. This time, let’s concentrate on queens. As you probably know, the queens may move not only

horizontally and vertically, but also diagonally.

You are given a chessboard with i−1 queens already placed and your task is to find all squares that may be used to place the i-th queen such that it cannot be captured by any of the others.

Input

The input consists of several tasks. Each task begins with a line containing three integer numbers separated by a space: X, Y , N. Xand Y give the chessboard size, 1 ≤ X, Y ≤20 000. N = i−1 is the number of queens already placed, 0 ≤ N ≤ X·Y .

After the first line, there are N lines, each containing two numbers xk, yk separated by a space. They give the position of the k-th queen, 1 ≤ xk ≤ X, 1 ≤ yk ≤ Y . You may assume that those positions are distinct, i.e., no two queens share the same square.

The last task is followed by a line containing three zeros.

Output

For each task, output one line containing a single integer number: the number of squares which are not occupied and do not lie on the same row, column, or diagonal as any of the existing queens.

Sample Input

Sample Output

题意：给出X*Y的棋盘，给出n个皇后求在棋盘上还有多少位置可以放置皇后；

分析：标记被占据位置的横行纵行，正副对角线对应的位置，for循环一次找出未被标记的点，（被标记的点可能被多次标记过，不容易计算计算未被标记的更方便）

code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std; 
int c[31000],l[31000],fd[81000],zd[81000];
int main()
{
    int i,j,xt,yt,X,Y,N;
    long long ans;
    while(scanf("%d%d%d",&X,&Y,&N))
    {
        if(X==0&&Y==0&&N==0)
            break;
        memset(c,0,sizeof(c));
        memset(l,0,sizeof(l));
        memset(fd,0,sizeof(fd));
        memset(zd,0,sizeof(zd));
        for(i=0;i<N;i++)
        {
            scanf("%d%d",&xt,&yt);
            c[xt]=1;
            l[yt]=1;
            fd[xt+yt]=1;
            zd[xt+Y-yt]=1;
        }
        ans=0;
        for(i=1;i<=X;i++)
            for(j=1;j<=Y;j++)
            {
                if(!c[i] && !l[j] && !fd[i+j] && !zd[i+Y-j])
                ans++;
            }
        cout<<ans<<endl;
    }
    return 0;
}