今天的D题YY了好久，能想到的情况都过了。最后说是网络流
签到：A. Lawn mower
表示水了，没想到转换，囧，就是把横和竖分别判一下就可以了。判断相邻两个点的距离是不是超过val，对于起点
和末点分别判断val/2.0就可以了：
代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int QuickMod(int a,int b,int n)
{
int r = 1;
while(b)
{
if(b&1)
r = (r*a)%n;
a = (a*a)%n;
b >>= 1;
}
return r;
}
#define maxn 1005
double X[maxn], Y[maxn];
int f1, f2;
int main()
{
int n, m;
double val;
while(scanf("%d%d%lf", &n, &m, &val))
{
if(!n && !m && val == 0.0)break;
for(int i = 1; i <= n; ++i)scanf("%lf", &X[i]);
for(int j = 1; j <= m; ++j)scanf("%lf", &Y[j]);
X[0] = 0;
X[n+1] = 75;
Y[0] = 0;
Y[m+1] = 100;
f1 = 0, f2 = 0;
sort(X, X + 1 + n);
sort(Y, Y + 1 + m);
if(X[1] - X[0] > val/2.0 || X[n+1] - X[n] > val/2.0)f1 = 1;
for(int i = 1; i < n; ++i)
if(X[i+1] - X[i] > val)f1 = 1;
if(Y[1] - Y[0] > val/2.0 || Y[m+1] - Y[m] > val/2.0)f2 = 1;
for(int i = 1; i < m; ++i)
if(Y[i+1] - Y[i] > val)f2 = 1;
if(!f1 && !f2)printf("YES\n");
else printf("NO\n");
}
return 0;
}