VK Cup 2015 - Round 1 E. Rooks and Rectangles 线段树 定点修改,区间最小值
E. Rooks and Rectangles
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/problemset/problem/524/E
Description
Input
Output
Sample Input
1 1
3 2
2 3
2 3 2 3
2 1 3 3
1 2 2 3
Sample Output
YES
NO
HINT
Picture to the sample: For the last area the answer is "NO", because cell (1, 2) cannot be hit by a rook.
题意
题解:
维护两个数据结构分别表示前i行里第几列是否被覆盖到和前i列里第j行是否被覆盖到
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 500001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
inline ll read()
{
int 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 buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
printf("\n");
}
*/
//**************************************************************************************
int a[maxn],n,m,k,q;;
struct node
{
int x,y;
};
node p[maxn];
struct pp
{
int x1,y1,x2,y2,id;
};
pp que[maxn]; void updata(int xx, int L, int R, int x, int val)
{
if(L==R)
{
a[xx]=val;
return;
}
int M=(L+R)>>;
if(x <= M)
updata(xx*,L,M,x,val);
else
updata(xx*+,M+,R,x,val);
a[xx]=min(a[xx*], a[xx*+]);
}
int query(int x, int L, int R, int l, int r)
{
if(l<=L&&R<=r)
return a[x];
int M=(L+R)>>;
if(r<=M)
return query(x*,L,M,l,r);
else if(l>M)
return query(x*+,M+,R,l,r);
else
return min(query(x*,L,M,l,r), query(x*+,M+,R,l,r));
}
bool cmp(node x,node y)
{
return x.x<y.x;
}
bool cmp1(pp x,pp y)
{
return x.x2<y.x2;
}
int ans[maxn];
void solve()
{
memset(a,,sizeof(a));
int pic=;
for(int i=;i<q;i++)
{
while(pic<k&&p[pic].x<=que[i].x2)
{
updata(,,m,p[pic].y,p[pic].x);
pic++;
}
if(query(,,m,que[i].y1,que[i].y2)>=que[i].x1)
ans[que[i].id]=;
}
}
void change()
{
swap(n,m);
for(int i=;i<k;i++)
swap(p[i].x,p[i].y);
sort(p,p+k,cmp);
for(int i=;i<q;i++)
{
swap(que[i].x1,que[i].y1);
swap(que[i].x2,que[i].y2);
}
sort(que,que+q,cmp1);
}
int main()
{ scanf("%d%d%d%d",&n,&m,&k,&q);
for(int i=;i<k;i++)
scanf("%d%d",&p[i].x,&p[i].y);
sort(p,p+k,cmp);
for(int i=;i<q;i++)
{
scanf("%d%d%d%d",&que[i].x1,&que[i].y1,&que[i].x2,&que[i].y2);
que[i].id=i;
}
sort(que,que+q,cmp1);
solve();
change();
solve();
for(int i=;i<q;i++)
{
if(ans[i])
puts("YES");
else
puts("NO");
}
}
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