Codeforces701B Cells Not Under Attack 数学推理

最新推荐文章于 2020-08-08 21:44:54 发布

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本文介绍了一个关于棋盘上放置车的问题，通过分析车的攻击范围来计算未受攻击格子的数量。文章提供了详细的解题思路及代码实现，适用于算法初学者理解如何利用数据结构解决实际问题。

Vasya has the square chessboard of size n × n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.

The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.

You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000,1 ≤ m ≤ min(100 000, n2)) — the size of the board and the number of rooks.

Each of the next m lines contains integers xi and yi (1 ≤ xi, yi ≤ n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.

Output

Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first i rooks are put.

   Examples
 

     input
   
3 3
1 1
3 1
2 2

     output
   
4 2 0 

     input
   
5 2
1 5
5 1

     output
   
16 9 

     input
   
100000 1
300 400

     output
   
9999800001 

Note

On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.

题意是放棋子，没有与任何棋子处于同一行的格子数为多少。
观察样例，可以发现一个事实，剩下的格子如果还存在的话，一定能够成一个矩形，我们从这个角度去考虑；

看样例，放了第一个棋子，长宽都小了一，那是不是每放一个棋子，长宽均减一呢，并不是这样的，第二个只是宽度减一；

从这，我们可以看出后放的棋子只有不与先前的棋子同列，宽才减一；不与先前的棋子同行，长才减一；
可用 v 数组记录前面出现过的棋子的横纵坐标。

两个注意点：答案超 int ；长宽等于0就直接输出0.

#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
int x[100005],y[100005];
bool v1[100005],v2[100005];
int main()
{
    int m;
    __int64 n,p,q;
    while(~scanf("%I64d%d",&n,&m)){
    p=q=n-1;
    for(int i=0;i<m;i++){
        scanf("%d%d",&x[i],&y[i]);
    }
    memset(v1,false,sizeof(v1));
    memset(v2,false,sizeof(v2));
    for(int t=0;t<m;t++){
        if(t==0){
            v1[x[0]]=true;
            v2[y[0]]=true;
            printf("%I64d",p*q);
        }else{
            if(!p||!q){
                printf(" 0");
            }else{
                if(!v1[x[t]]){
                    p--;
                }
                if(!v2[y[t]]){
                    q--;
                }
                v1[x[t]]=v2[y[t]]=true;
                printf(" %I64d",p*q);
            }
        }
    }
    printf("\n");
    }
    return 0;
}