Codeforces Round #587 (Div. 3)C--White Sheet（矩形交，三种解法）
http://codeforces.com/contest/1216/problem/C
题意:白色矩形是否被黑色矩形完全覆盖，是输出NO
解法:
如果只说对角线没说明是哪条对角线：
if(a1.x1>a1.x2)swap(a1.x1,a1.x2);
if(a1.y1>a1.y2)swap(a1.y1,a1.y2);
if(a2.x1>a2.x2)swap(a2.x1,a2.x2);
if(a2.y1>a2.y2)swap(a2.y1,a2.y2);
一.相交面积==白色
这做法看起来思路简单但是很多坑
应该是两个黑白相交面积之和-黑黑白相交面积，而不是-黑黑相交面积！(当时就wa在这)
求相交面积先判断是否相交（即x1,x2大小与y1,y2大小）
#include
using namespace std;
typedef long long ll;
struct c1{
ll x1,y1,x2,y2;
};
c1 jiao(c1 a,c1 b){//相交矩形坐标（可能不存在）
c1 c;
c.x1 = max(a.x1, b.x1);
c.y1 = max(a.y1, b.y1);
c.x2 = min(a.x2, b.x2);
c.y2 = min(a.y2, b.y2);
return c;
}
int main(){
c1 a1,a2,b;
cin>>b.x1>>b.y1>>b.x2>>b.y2>>a1.x1>>a1.y1>>a1.x2>>a1.y2>>a2.x1>>a2.y1>>a2.x2>>a2.y2;
c1 j1=jiao(b,a1);
c1 j2=jiao(b,a2);
c1 j3=jiao(j1,j2);
ll s,s1,s2,s3;
if(b.x1>b.x2||b.y1>b.y2)s=0;//特判！
else s=(b.x2-b.x1)*(b.y2-b.y1);
if(j1.x1>j1.x2||j1.y1>j1.y2)s1=0;
else s1=(j1.x2-j1.x1)*(j1.y2-j1.y1);
if(j2.x1>j2.x2||j2.y1>j2.y2)s2=0;
else s2=(j2.x2-j2.x1)*(j2.y2-j2.y1);
if(j3.x1>j3.x2||j3.y1>j3.y2)s3=0;
else s3=(j3.x2-j3.x1)*(j3.y2-j3.y1);
if(s1+s2-s3==s)cout<<"NO"<
二.分类讨论
根据黑白相对方位，上下左右分别判断
#include
using namespace std;
struct node
{
int x1,x2,y1,y2;
}a,b,c;
int main()
{
cin>>a.x1>>a.y1>>a.x2>>a.y2>>b.x1>>b.y1>>b.x2>>b.y2>>c.x1>>c.y1>>c.x2>>c.y2;
if(c.x1<=a.x1&&c.x2>=a.x2 &&c.y1<=a.y1&&c.y2>=a.y2 ) puts("NO");
else if(b.x1<=a.x1&&b.x2>=a.x2 &&b.y1<=a.y1&&b.y2>=a.y2 ) puts("NO");
else if(c.x1<=a.x1&&c.x2>=a.x2&&b.x1<=a.x1&&b.x2>=a.x2&&b.y1<=a.y1&&c.y2>=a.y2&&b.y2>=c.y1) puts("NO");
else if(c.x1<=a.x1&&c.x2>=a.x2&&b.x1<=a.x1&&b.x2>=a.x2&&c.y1<=a.y1&&b.y2>=a.y2&&c.y2>=b.y1) puts("NO");
else if(c.y1<=a.y1&&c.y2>=a.y2&&b.y1<=a.y1&&b.y2>=a.y2&&b.x1<=a.x1&&c.x2>=a.x2&&b.x2>=c.x1) puts("NO");
else if(c.y1<=a.y1&&c.y2>=a.y2&&b.y1<=a.y1&&b.y2>=a.y2&&c.x1<=a.x1&&b.x2>=a.x2&&c.x2>=b.x1) puts("NO");
else puts("YES");
}
三.离散化➕二维数组vis覆盖
注意有种情况，只覆盖相邻两个整点，剩下中间不是整点所以*2，*2+-1代表附近半格
#include
#define maxl 300010
using namespace std;
int n,m,ans,cntx,cnty,totx,toty;
int x[7],y[7];
int numx[200],numy[200];
int a[200][200];
char s[maxl];
inline void prework()
{
for(int i=1;i<=6;i++)
{
scanf("%d%d",&x[i],&y[i]);
x[i]*=2;y[i]*=2;
numx[++cntx]=x[i];
numx[++cntx]=x[i]-1;
numx[++cntx]=x[i]+1;
numy[++cnty]=y[i];
numy[++cnty]=y[i]-1;
numy[++cnty]=y[i]+1;
}
sort(numx+1,numx+1+cntx);
sort(numy+1,numy+1+cnty);
totx=unique(numx+1,numx+1+cntx)-numx-1;
toty=unique(numy+1,numy+1+cnty)-numy-1;
}
inline void mainwork()
{
int x1,x2,y1,y2;
x1=lower_bound(numx+1,numx+1+totx,x[3])-numx;
x2=lower_bound(numx+1,numx+1+totx,x[4])-numx;
y1=lower_bound(numy+1,numy+1+toty,y[3])-numy;
y2=lower_bound(numy+1,numy+1+toty,y[4])-numy;
for(int i=x1;i<=x2;i++)
for(int j=y1;j<=y2;j++)
a[i][j]=1;
x1=lower_bound(numx+1,numx+1+totx,x[5])-numx;
x2=lower_bound(numx+1,numx+1+totx,x[6])-numx;
y1=lower_bound(numy+1,numy+1+toty,y[5])-numy;
y2=lower_bound(numy+1,numy+1+toty,y[6])-numy;
for(int i=x1;i<=x2;i++)
for(int j=y1;j<=y2;j++)
a[i][j]=1;
ans=0;
x1=lower_bound(numx+1,numx+1+totx,x[1])-numx;
x2=lower_bound(numx+1,numx+1+totx,x[2])-numx;
y1=lower_bound(numy+1,numy+1+toty,y[1])-numy;
y2=lower_bound(numy+1,numy+1+toty,y[2])-numy;
for(int i=x1;i<=x2;i++)
for(int j=y1;j<=y2;j++)
if(a[i][j]==0)
ans=1;
}
inline void print()
{
if(ans>0)
puts("YES");
else
puts("NO");
}
int main()
{
int t=1;
//scanf("%d",&t);
for(int i=1;i<=t;i++)
{
prework();
mainwork();
print();
}
return 0;
}