[原创]Codeforces Round #364 div.2 B 【数组标记】

2016-07-23 11:07:59 Tabris_ 阅读数：442

博客爬取于2020-06-14 22:44:09
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52003146

题目连接 ：http://codeforces.com/problemset/problem/701/B

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B. Cells Not Under Attack
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
这里写图片描述
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题目大意： 就是在N*N的方格内 放入M个 “車” 这样 这个"車"所在的行列都是不安全的了 问 一次放入这些“車”后 还有多少安全的位置
（自己大概意淫出来的正确的题意）

题解 ： N，M 的数据量太大了，不能暴力求解 。 xjb画了画 发现 每次安全的位置个数都是NN-N(x+y)+xy; （x表示有多少行上有车，y同理）最后再加上x和y减重的那部分xy；既是正解；

操作的时候要用两个数组记录下这行有没有車，这列有没有車 如果没有x，y就++；有了就不加了

附本体代码

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int xx[101010],yy[101010];

int main()
{
    LL n,m;
    scanf("%I64d%I64d",&n,&m);

    for(int i=0;i<=n;i++)
        xx[i]=0,yy[i]=0;

    LL x,y,xxx=0,yyy=0;

    for(int i=0;i<m;i++)
    {
        scanf("%I64d%I64d",&x,&y);
        if(!xx[x]) xxx++,xx[x]=1;
        if(!yy[y]) yyy++,yy[y]=1;
        printf("%I64d ",n*n-n*(xxx+yyy)+xxx*yyy);
    }
    puts("");

    return 0;
}