Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100187/problem/F
Description
Doomsday comes in t units of time. In anticipation of such a significant event n people prepared m vaults in which, as they think, it will be possible to survive. But each vault can accommodate only k people and each person can pass only one unit of distance per one unit of time. Fortunately, all people and vaults are now on the straight line, so there is no confusion and calculations should be simple.
You are given the positions of the people and the vaults on the line. You are to find the maximal number of people who can hide in vaults and think they will survive.
Input
The first line contains four integers n, m, k and t (1 ≤ n, m, k ≤ 200000, 1 ≤ t ≤ 109) separated by spaces — the number of people, the number of vaults, the capacity of one vault and the time left to the Doomsday.
The second line contains n integers separated by spaces — the coordinates of the people on the line.
The third line contains m integers separated by spaces — the coordinates of the vaults on the line.
All the coordinates are between - 109 and 109, inclusively.
Output
Output one integer — the maximal number of people who can hide in vaults and think they will survive.
Sample Input
2 2 1 5
45 55
40 60
Sample Output
2
HINT
题意
有n个人,m个藏身处，每个藏身处可以藏k个人，一个人可以往左边走t米，往右边走t米，问你最后能躲几个人？
题解：
这是一个区间覆盖问题，类似贪心搞一搞就行了
代码
#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> #include <stack> 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 test freopen("test.txt","r",stdin) #define maxn 200101 #define mod 1000000009 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll 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 node { int x,y; }; bool cmp(node a,node b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } ll flag[maxn]; node a[maxn]; ll b[maxn]; ll c[maxn]; int main() { ll n=read(),m=read(),k=read(),t=read(); for(int i=0;i<n;i++) { int x=read(); a[i].x=x-t; a[i].y=x+t; } for(int j=0;j<m;j++) b[j]=read(); sort(a,a+n,cmp); sort(b,b+m); int ans=0; int j=0; for(int i=0;i<n;i++) { if(j==m) break; while(b[j]<a[i].x&&j<m) j++; if(j==m) break; if(a[i].y>=b[j]) { ans++; c[j]++; if(c[j]==k) j++; } } cout<<ans<<endl; }
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100187/problem/F
Description
Doomsday comes in t units of time. In anticipation of such a significant event n people prepared m vaults in which, as they think, it will be possible to survive. But each vault can accommodate only k people and each person can pass only one unit of distance per one unit of time. Fortunately, all people and vaults are now on the straight line, so there is no confusion and calculations should be simple.
You are given the positions of the people and the vaults on the line. You are to find the maximal number of people who can hide in vaults and think they will survive.
Input
The first line contains four integers n, m, k and t (1 ≤ n, m, k ≤ 200000, 1 ≤ t ≤ 109) separated by spaces — the number of people, the number of vaults, the capacity of one vault and the time left to the Doomsday.
The second line contains n integers separated by spaces — the coordinates of the people on the line.
The third line contains m integers separated by spaces — the coordinates of the vaults on the line.
All the coordinates are between - 109 and 109, inclusively.
Output
Output one integer — the maximal number of people who can hide in vaults and think they will survive.
Sample Input
2 2 1 5
45 55
40 60
Sample Output
2
HINT
题意
有n个人,m个藏身处，每个藏身处可以藏k个人，一个人可以往左边走t米，往右边走t米，问你最后能躲几个人？
题解：
这是一个区间覆盖问题，类似贪心搞一搞就行了
代码
#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> #include <stack> 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 test freopen("test.txt","r",stdin) #define maxn 200101 #define mod 1000000009 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll 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 node { int x,y; }; bool cmp(node a,node b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } ll flag[maxn]; node a[maxn]; ll b[maxn]; ll c[maxn]; int main() { ll n=read(),m=read(),k=read(),t=read(); for(int i=0;i<n;i++) { int x=read(); a[i].x=x-t; a[i].y=x+t; } for(int j=0;j<m;j++) b[j]=read(); sort(a,a+n,cmp); sort(b,b+m); int ans=0; int j=0; for(int i=0;i<n;i++) { if(j==m) break; while(b[j]<a[i].x&&j<m) j++; if(j==m) break; if(a[i].y>=b[j]) { ans++; c[j]++; if(c[j]==k) j++; } } cout<<ans<<endl; }
