今日SGU 6.6

sgu 177

题意：给你一个一开始全是白色的正方形，边长为n，然后问你经过几次染色之后，最后的矩形里面

还剩多少个白色的块

收获：矩形切割，我们可以这么做，离散处理，对于每次染黑的操作，看看后面有没有染白的矩形和它这个染黑的

矩形有相交，那么我们就把它重合的部分减掉

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int N = 5e3+6;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
int x[N],y[N],y2[N],x2[N],col[N];
int ans;
void gao(int d, int xs, int ys, int xe, int ye) {
    while(d <= m && ( xs > x2[d] || xe < x[d] || ys > y2[d] || ye < y[d] ) ) d++;
    if(d == m + 1) {
        ans -= (ye - ys + 1) * (xe - xs + 1);
        return ;
    }
    // 想想被每个后面的矩形的四条边，每条切割完后的剩下什么 
    int k1 = max(xs, x[d]);
    if(xs < k1) gao(d + 1, xs, ys, k1 - 1, ye);
    int k2 = min(xe, x2[d]);
    if(xe > k2) gao(d + 1, k2 + 1, ys, xe, ye);
    xs = k1, xe = k2;
    k1 = max(ys, y[d]);
    if(ys < k1) gao(d + 1, xs, ys, xe, k1 - 1);
    k2 = min(ye, y2[d]);
    if(ye > k2) gao(d + 1, xs, k2 + 1, xe, ye); 
}
int main(){
    scanf("%d%d",&n,&m); ans = n * n;
    rep(i, 1, m+1) {
        char c;
        scanf("%d%d%d%d %c",&x[i],&y[i],&x2[i],&y2[i],&c);
        if(x[i] > x2[i]) swap(x2[i], x[i]);
        if(y[i] > y2[i]) swap(y2[i], y[i]); 
        col[i] = c == 'b' ? 1 : 0;
    }
    repd(i,m,1) if(col[i]) gao(i+1,x[i],y[i],x2[i],y2[i]);
    printf("%d
",ans);
    return 0;
}