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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 arenot under attack after Vasya puts it on the board.
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.
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.
3 3
1 1
3 1
2 2
4 2 0
5 2
1 5
5 1
16 9
100000 1
300 400
9999800001
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.
分析：模拟即可；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #include <ext/rope> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define vi vector<int> #define pii pair<int,int> #define mod 1000000007 #define inf 0x3f3f3f3f #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) const int maxn=1e5+10; const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}}; using namespace std; using namespace __gnu_cxx; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,x,y,cnt1,cnt2; bool a[maxn],b[maxn]; int main() { int i,j,k,t; scanf("%d%d",&n,&m); ll now=1ll*n*n; while(m--) { scanf("%d%d",&x,&y); if(!a[x]) { now-=n; now+=cnt2; } if(!b[y]) { now-=n; now+=cnt1; } if(!a[x])cnt1++; if(!b[y])cnt2++; if(!a[x]&&!b[y])now++; a[x]=b[y]=true; printf("%lld ",now); } //system ("pause"); return 0; }
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原文地址：http://www.cnblogs.com/dyzll/p/5700622.html
