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.
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.
If it is possible to distribute tickets between people in the described manner print "YES", otherwise print "NO".
2 2
3 3 3 2
1 3
YES
2 2
3 2 3 3
1 2
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 }