2 seconds
256 megabytes
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 lintegers 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
题目大意：n*m个座位， 有n*m个人，一开始有k个人在(0,0)点上，l个人在(0，m+1)点上，每个人有对应的体力值，体力值即为可以行走的距离（曼哈顿距离），问是否存在一种方案是每个人花费的体力不超过上限，且每个人都有位置坐。
贪心：对于前k个人，我们按照体力排序，显然找到一个以距离（0，m+1)尽可能远为第一关键字，与（0,0）尽可能远为第二关键字的位置，那么这个人就应该在这个位置。之后l个人放到与（0，m+1)尽可能远的且没有人的位置，检测是否存在这种可能。
AC code
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
#include<ctime>
#include<cstring>
#define yyj(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout);
#define llg long long
#define maxn 10010
using namespace std;
llg i,j,k,n,m,k1,k2,t1,t2,bj[maxn],x,y,l,len,a[maxn],b[maxn],maxl;
bool f;
int main()
{
yyj("");
cin>>n>>m>>k1;
for (i=;i<=k1;i++) cin>>a[i];
sort(a+,a+k1+);
cin>>k2;
for (j=;j<=k2;j++) cin>>b[j];
sort(b+,b+k2+);
int c[n+][m+];
for (i=;i<=n;i++) for (j=;j<=m;j++) c[i][j]=;
for (k=;k<=k1;k++)
{
f=false; maxl=;
for (i=;i<=n;i++)
for (j=;j<=m;j++)
if (i+j<=a[k] && i+m+-j>maxl && c[i][j]==)
{
f=true;
x=i,y=j;
maxl=i+m-j+;
}
if (f)
{
c[x][y]=;
}
else
{
cout<<"NO";
return ;
}
}
for (k=;k<=k2;k++)
{
maxl=,f=false;
for (i=;i<=n;i++)
for (j=;j<=m;j++)
if (c[i][j]== && i+m+-j>maxl && i+m+-j<=b[k])
{
f=true;
x=i; y=j;
maxl=i+m+-j;
}
if (f)
{
c[x][y]=;
}
else
{
cout<<"NO";
return ;
}
}
cout<<"YES";
return ;
}