bzoj1623 [Usaco2008 Open]Cow Cars 奶牛飞车
来源:互联网 发布:vba 递归算法 编辑:程序博客网 时间:2024/04/29 12:15
Description
编号为1到N的N只奶牛正各自驾着车打算在牛德比亚的高速公路上飞驰.高速公路有M(1≤M≤N)条车道.奶牛i有一个自己的车速上限Si(l≤Si≤1,000,000).
在经历过糟糕的驾驶事故之后,奶牛们变得十分小心,避免碰撞的发生.每条车道上,如果某一只奶牛i的前面有南只奶牛驾车行驶,那奶牛i的速度上限就会下降kD个单位,也就是说,她的速度不会超过Si - kD(O≤D≤5000),当然如果这个数是负的,那她的速度将是0.牛德比亚的高速会路法规定,在高速公路上行驶的车辆时速不得低于/(1≤L≤1,000,000).那么,请你计算有多少奶牛可以在高速公路上行驶呢?
Input
第1行输入N,M,D,L四个整数,之后N行每行一个整数输入Si.
Output
输出最多有多少奶牛可以在高速公路上行驶.
Sample Input
3 1 1 5//三头牛开车过一个通道.当一个牛进入通道时,它的速度V会变成V-D*X(X代表在它前面有多少牛),它减速后,速度不能小于L
5
7
5
INPUT DETAILS:
There are three cows with one lane to drive on, a speed decrease
of 1, and a minimum speed limit of 5.
5
7
5
INPUT DETAILS:
There are three cows with one lane to drive on, a speed decrease
of 1, and a minimum speed limit of 5.
Sample Output
2
OUTPUT DETAILS:
Two cows are possible, by putting either cow with speed 5 first and the cow
with speed 7 second.
OUTPUT DETAILS:
Two cows are possible, by putting either cow with speed 5 first and the cow
with speed 7 second.
看到那么多大神都做了这题……我也去写
贪心……排序一下显然速度小的要先合并
自然而然的想到了平衡树……(不要D我)
但是黄巨大说直接每次需按人数最少的加进去就好了
当然我这sillycross想不到这么精妙的做法
#include<cstdio>#include<algorithm>using namespace std;int v[50010];int n,m,d,l,forward,ans;inline int read(){ int 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;}int main(){n=read();m=read();d=read();l=read();for (int i=1;i<=n;i++) v[i]=read();sort(v+1,v+n+1);for (int i=1;i<=n;i++) { forward=ans/m; if (v[i]-forward*d>=l)ans++; }printf("%d",ans);}
0 0
- bzoj1623 [Usaco2008 Open]Cow Cars 奶牛飞车
- [BZOJ1623][Usaco2008 Open]Cow Cars 奶牛飞车
- bzoj1623【Usaco2008 Open】Cow Cars 奶牛飞车
- [Usaco2008 Open]Cow Cars 奶牛飞车
- BZOJ 1623: [Usaco2008 Open]Cow Cars 奶牛飞车
- 贪心-BZOJ-1623-[Usaco2008 Open]Cow Cars 奶牛飞车
- BZOJ 1623: [Usaco2008 Open]Cow Cars 奶牛飞车
- bzoj 1623: [Usaco2008 Open]Cow Cars 奶牛飞车(贪心)
- [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- BZOJ 1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- BZOJ 1604 [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居 Treap
- bzoj1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- 【bzoj1604】: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- [BZOJ1604] [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- bzoj 1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- BZOJ 1604 [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- 并查集+Set-BZOJ-1604-[Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
- 【bzoj 1604】: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居 set+并查集
- Java 泛型(super和extends关键字)
- 【转载】CString,string,char数组的转换
- win7+ubuntu 13.04双系统安装方法
- JTA事务总结
- POJ 1062 昂贵的聘礼 最短路
- bzoj1623 [Usaco2008 Open]Cow Cars 奶牛飞车
- 到底是用Margin还是用Padding(效果相同时)
- 最小二叉树
- HDU 1698-I Hate It (线段树,点修改)
- HDU 4883 TIANKENG’s restaurant Bestcoder 2-1(模拟)
- 俄方不舒服那个师傅给你发撒旦给你发个你过分
- 转:解释执行和编译执行的区别
- HDOJ 题目2035人见人爱A^B(剩余定理 水题)
- wscratchpad.js 插件在安卓系统上不兼容