hdu 5480 Conturbatio 线段树 单点更新，区间查询最小值
Conturbatio
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5480Description
There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.
There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?
Input
The first line of the input is a integer T, meaning that there are T test cases.
Every test cases begin with four integers n,m,K,Q.
K is the number of Rook, Q is the number of queries.
Then K lines follow, each contain two integers x,y describing the coordinate of Rook.
Then Q lines follow, each contain four integers x1,y1,x2,y2 describing the left-down and right-up coordinates of query.
1≤n,m,K,Q≤100,000.
1≤x≤n,1≤y≤m.
1≤x1≤x2≤n,1≤y1≤y2≤m.
Output
For every query output "Yes" or "No" as mentioned above.
Sample Input
2
2 2 1 2
1 1
1 1 1 2
2 1 2 2
2 2 2 1
1 1
1 2
2 1 2 2
Sample Output
Yes
No
Yes
HINT
题意
在一个n \times mn×m的国际象棋棋盘上有很多车(Rook)，其中车可以攻击他所属的一行或一列，包括它自己所在的位置。
现在还有很多询问，每次询问给定一个棋盘内部的矩形，问矩形内部的所有格子是否都被车攻击到？
题解：
我是线段树做的，只要统计x1,x2这个区域的最小值和y1 y2这个区域的最小值都不同时为0 就好了
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 150001 #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; } */ //************************************************************************************** struct data{ int l,r,mn; }tr[maxn*4]; data tr2[maxn*4]; void build2(int k,int s,int t) { tr2[k].l=s;tr2[k].r=t; if(s==t){tr2[k].mn=0;return;} int mid=(s+t)>>1; build2(k<<1,s,mid); build2(k<<1|1,mid+1,t); tr2[k].mn=min(tr2[k<<1].mn,tr2[k<<1|1].mn); } int ask2(int k,int s,int t) { int l=tr2[k].l,r=tr2[k].r; if(s==l&&t==r)return tr2[k].mn; int mid=(l+r)>>1; if(t<=mid)return ask2(k<<1,s,t); if(s>mid)return ask2(k<<1|1,s,t); return min(ask2(k<<1,s,mid),ask2(k<<1|1,mid+1,t)); } void update2(int k,int x,int y) { int l=tr2[k].l,r=tr2[k].r; if(l==r){tr2[k].mn=y;return;} int mid=(l+r)>>1; if(x<=mid)update2(k<<1,x,y); if(x>mid)update2(k<<1|1,x,y); tr2[k].mn=min(tr2[k<<1].mn,tr2[k<<1|1].mn); } void build(int k,int s,int t) { tr[k].l=s;tr[k].r=t; if(s==t){tr[k].mn=0;return;} int mid=(s+t)>>1; build(k<<1,s,mid); build(k<<1|1,mid+1,t); tr[k].mn=min(tr[k<<1].mn,tr[k<<1|1].mn); } int ask(int k,int s,int t) { int l=tr[k].l,r=tr[k].r; if(s==l&&t==r)return tr[k].mn; int mid=(l+r)>>1; if(t<=mid)return ask(k<<1,s,t); if(s>mid)return ask(k<<1|1,s,t); return min(ask(k<<1,s,mid),ask(k<<1|1,mid+1,t)); } void update(int k,int x,int y) { int l=tr[k].l,r=tr[k].r; if(l==r){tr[k].mn=y;return;} int mid=(l+r)>>1; if(x<=mid)update(k<<1,x,y); if(x>mid)update(k<<1|1,x,y); tr[k].mn=min(tr[k<<1].mn,tr[k<<1|1].mn); } int main() { int t;scanf("%d",&t); while(t--) { int n,m,k,q; scanf("%d%d%d%d",&n,&m,&k,&q); build(1,1,n); build2(1,1,m); for(int i=1;i<=k;i++) { int x,y; scanf("%d%d",&x,&y); update(1,x,1); update2(1,y,1); } for(int i=1;i<=q;i++) { int x1,x2,y1,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); if(x1>x2)swap(x1,x2); if(y1>y2)swap(y1,y2); int d1 = max(ask(1,x1,x2),ask2(1,y1,y2)); if(d1==1) printf("Yes\n"); else printf("No\n"); } } return 0; }
