扫描线 - UVALive - 6864 Strange Antennas

Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=87213

[b]Mean:[/b]

给你一个N*N的网格，有M个雷达，每个雷达的扫射区域是一个直角边长为P的等腰直角三角形，能够向以直角顶点为中心的四个象限扫射。

雷达之间存在信号屏蔽，只有被奇数个雷达扫射到的区域才能被信号覆盖。求被信号覆盖的区域是多少。

[b]analyse:[/b]

因为给的都是整数点，这样就不涉及到计算几何了。

N的范围是0~3000，M的范围是1~100。

总的思路是：对N*N的网格做一维遍历（随便哪一维都行）。假设我们选取横向遍历，那么对于每一个竖列，我们对所有的雷达进行判断，判断是否经过这个数列。

如果经过，计算出在这个数列的长度，用一个数组存起来。每扫一列，对这一列的线段做一次扫描线，加到answer中，扫完即得最终答案。

[b]Time complexity: O(N*M)[/b]

[b]Source code: [/b]

代码1：

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-08-22.04
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;

const int MAXN = 30005,MAXM = 105;
void scan(int &x)
{
x=0;
char c=getchar();
while(!(c>='0' && c<='9' || c=='-')) { c=getchar(); }
bool flag=1;
if(c=='-')
{
flag=0; c=getchar();
}
while(c>='0' && c<='9')
{
x=x*10+c-'0'; c=getchar();
}
if(!flag) { x=-x; }
}
void scan2(int &x,int &y) { scan(x),scan(y);}
void scan3(int &x,int &y,int &z) { scan(x),scan(y),scan(z); }
/**************************************END define***************************************/
struct Event
{
int x, val;
Event() {}
Event(int _x, int _val) : x(_x), val(_val) {}
bool operator < (const Event &a) const
{
return x < a.x || (x == a.x && val < a.val);
}
};

int n, m;
int x[MAXM],y[MAXM],p[MAXM],d[MAXM];
char str[10];

Event event[MAXM * 2];

int main()
{
while(scanf("%d", &n) != EOF)
{
scan(m);
for(int i = 0; i < m; ++i)
{
scan2(x[i],y[i]),scan2(p[i],d[i]);
if(d[i] == 0) --y[i];
if(d[i] == 2) --x[i];
if(d[i] == 3) --x[i],--y[i];
}
int ans = 0;
for(int i = 0; i < n; ++i)
{
int cntEvent = 0;
for(int j = 0; j < m; ++j)
{
int low = -1, high = -1,len;
if(d[j] == 0 && x[j] <= i && i <= x[j] + p[j] - 1)
{
len = p[j] - (i - x[j]);
low = max(y[j] - len + 1, 0);
high = y[j];
}
if(d[j] == 1 && x[j] <= i && i <= x[j] + p[j] - 1)
{
len = p[j] - (i - x[j]);
low = y[j];
high = min(y[j] + len - 1, n - 1);
}
if(d[j] == 2 && x[j] - p[j] + 1 <= i && i <= x[j])
{
len = p[j] - (x[j] - i);
low = y[j];
high = min(y[j] + len - 1, n - 1);
}
if(d[j] == 3 && x[j] - p[j] + 1 <= i && i <= x[j])
{
len = p[j] - (x[j] - i);
low = max(y[j] - len + 1, 0);
high = y[j];
}
if(low != -1 && high != -1 && low <= high)
{
// printf( "%d %d\n", low, high );
event[cntEvent++] = Event(low, 1);
event[cntEvent++] = Event(high + 1, -1);
}
}
sort(event, event + cntEvent);
int res = 0,sum = 0;
for(int j = 0; j < cntEvent; ++j)
{
if((sum & 1))
res += event[j].x - event[j - 1].x;
sum += event[j].val;
}
// printf("%d\n", res);
ans += res;
}
printf("%d\n", ans);
}
return 0;
}
代码2：

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-08-13.52
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN=205;
int n,m;

void scan(int &x)
{
x=0;
char c=getchar();
while(!(c>='0' && c<='9' || c=='-')) { c=getchar(); }
bool pos=1;
if(c=='-')
{
pos=0; c=getchar();
}
while(c>='0' && c<='9')
{
x=x*10+c-'0'; c=getchar();
}
if(!pos) { x=-x; }
}
void scan2(int &x,int &y) { scan(x),scan(y);}
void scan3(int &x,int &y,int &z) { scan(x),scan(y),scan(z); }
/**************************************END define***************************************/

struct Tra
{
int x,y,p,k;
int x1,x2,y1,y2;
int xd,yd;
}d[MAXN];

struct Line
{
int pos,flag;
} Li[2*MAXN];
int cnt;
bool cmp(Line a,Line b)
{
return a.pos<b.pos;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
while(~scanf("%d",&m))
{
scan(n);
for(int i=1;i<=n;++i)
{
int x,y,p,k;
scan(x),scan(y),scan(p),scan(k);
d[i].x=x,d[i].y=y,d[i].p=p,d[i].k=k;
if(k==0)
{
d[i].x1=x,d[i].x2=x+p;
d[i].y1=y-p,d[i].y2=y;
d[i].xd=x+p,d[i].yd=y;
}
else if(k==1)
{
d[i].x1=x,d[i].x2=x+p;
d[i].y1=y,d[i].y2=y+p;
d[i].xd=x+p,d[i].yd=y;
}
else if(k==2)
{
d[i].x1=x,d[i].x2=x-p;
d[i].y1=y,d[i].y2=y+p;
d[i].xd=x,d[i].yd=y+p;
}
else
{
d[i].x1=x,d[i].x2=x-p;
d[i].y1=y-p,d[i].y2=y;
d[i].xd=x,d[i].yd=y-p;
}
}
int ans=0;
for(int yi=0;yi<m;++yi)
{
int from ,to ,t,xt;
cnt=0;
for(int i=1;i<=n;++i)
{
if(d[i].y1<=yi&&yi+1<=d[i].y2)
{
if(d[i].k<=1) from=to=d[i].x>d[i].x2?d[i].x2:d[i].x;
else from=to=d[i].x1>d[i].x2?d[i].x1:d[i].x2;

t=abs(d[i].yd-yi);
xt=d[i].xd-t;
from=from<xt?from:xt;
to=to>xt?to:xt;

t=abs(d[i].yd-(yi+1));
xt=d[i].xd-t;
from=from<xt?from:xt;
to=to>xt?to:xt;

from=from>0?from:0;
from++;
to=to<m?to:m;
cnt++;
Li[cnt].pos=from;
Li[cnt].flag=+1;

cnt++;
Li[cnt].pos=to+1;
Li[cnt].flag=-1;
}
}
cnt++;
Li[cnt].pos=m+1;
sort(Li+1,Li+1+cnt,cmp);
int lastpos=1,num=0;
for(int i=1;i<=cnt;)
{
if(num&1) ans+=Li[i].pos-lastpos;
lastpos=Li[i].pos;
int ti=i;
while(ti<=cnt&&Li[ti].pos==Li[i].pos)
{
num+=Li[ti].flag;
ti++;
}
i=ti;
}
}
printf("%d\n",ans);
}
return 0;
}
/*

*/