Description
The closing ceremony of Squanch Code Cup is held in the big hall with n × m seats, arranged in n rows, m seats in a row. Each seat has two coordinates (x, y) (1 ≤ x ≤ n, 1 ≤ y ≤ m).
There are two queues of people waiting to enter the hall: k people are standing at (0, 0) and n·m - k people are standing at (0, m + 1). Each person should have a ticket for a specific seat. If person p at (x, y) has ticket for seat (xp, yp) then he should walk |x - xp| + |y - yp| to get to his seat.
Each person has a stamina — the maximum distance, that the person agrees to walk. You should find out if this is possible to distribute all n·m tickets in such a way that each person has enough stamina to get to their seat.
Input
The first line of input contains two integers n and m (1 ≤ n·m ≤ 104) — the size of the hall.
The second line contains several integers. The first integer k (0 ≤ k ≤ n·m) — the number of people at (0, 0). The following k integers indicate stamina of each person there.
The third line also contains several integers. The first integer l (l = n·m - k) — the number of people at (0, m + 1). The following l integers indicate stamina of each person there.
The stamina of the person is a positive integer less that or equal to n + m.
Output
If it is possible to distribute tickets between people in the described manner print "YES", otherwise print "NO".
Examples
Input
2 23 3 3 21 3
Output
YES
Input
2 23 2 3 31 2
Output
NO 正解：贪心解题报告：
因为我们想使得到两个出发点的距离小，但是不能同时保证两个，那么我们只能先保证一个最优，才想办法判断另外一个。显然把所有点按到左下角距离排序，可以对于左下角的点判断可达，然后我们每次走离左上角尽可能远的点，这样相当于是帮左上角的点分担了一部分远的点，同时可以保证合法性。
1 //It is made by jump~
2 #include <iostream>
3 #include <cstdlib>
4 #include <cstring>
5 #include <cstdio>
6 #include <cmath>
7 #include <algorithm>
8 #include <ctime>
9 #include <vector>
10 #include <queue>
11 #include <map>
12 #include <set>
13 using namespace std;
14 typedef long long LL;
15 const int MAXN = 10011;
16 int n,m,k,tot;
17 int a[MAXN];
18 struct node{
19 int x,y,z,dis;
20 bool operator < (const node &a) const{
21 return a.z>z;
22 }
23 }s[MAXN];
24
25 priority_queue<node>Q;
26 inline int getint()
27 {
28 int w=0,q=0; char c=getchar();
29 while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); if(c==‘-‘) q=1,c=getchar();
30 while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w;
31 }
32
33 inline bool cmp(node q,node qq){ return q.dis<qq.dis; }
34
35 inline void work(){
36 n=getint(); m=getint(); k=getint();
37 for(int i=1;i<=k;i++) a[i]=getint();
38 sort(a+1,a+k+1);
39 for(int i=1;i<=n;i++)
40 for(int j=1;j<=m;j++)
41 s[++tot].x=i,s[tot].y=j,s[tot].z=m+1-j+i,s[tot].dis=i+j;
42 sort(s+1,s+tot+1,cmp); int u=1; bool ok=true;
43
44 for(int i=1;i<=k;i++) {
45 while(a[i]>=s[u].dis && u<=tot) Q.push(s[u]),u++;
46 if(Q.empty()) { ok=false; break; }
47 Q.pop();//清空
48 }
49 if(!ok) { printf("NO"); return; }
50 while(u<=tot) Q.push(s[u]),u++;
51
52 k=getint(); for(int i=1;i<=k;i++) a[i]=getint();
53 sort(a+1,a+k+1);
54 for(int i=k;i>=1;i--) {
55 if(a[i]<Q.top().z) { ok=false; break; }
56 Q.pop();
57 }
58 if(!ok) { printf("NO"); return; }
59 printf("YES");
60 }
61
62 int main()
63 {
64 work();
65 return 0;
66 }