Frog - HDU 5037 想法题

Frog

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 413    Accepted Submission(s): 89

Problem Description

Once upon a time, there is a little frog called Matt. One day, he came to a river.

The river could be considered as an axis.Matt is standing on the left bank now (at position 0). He wants to cross the river, reach the right bank (at position M). But Matt could only jump for at most L units, for example from 0 to L.

As the God of Nature, you must save this poor frog.There are N rocks lying in the river initially. The size of the rock is negligible. So it can be indicated by a point in the axis. Matt can jump to or from a rock as well as the bank.

You don't want to make the things that easy. So you will put some new rocks into the river such that Matt could jump over the river in maximal steps.And you don't care the number of rocks you add since you are the God.

Note that Matt is so clever that he always choose the optimal way after you put down all the rocks.

 

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains N, M, L (0<=N<=2*10^5,1<=M<=10^9, 1<=L<=10^9).

And in the following N lines, each line contains one integer within (0, M) indicating the position of rock.

 

Output

For each test case, just output one line “Case #x: y", where x is the case number (starting from 1) and y is the maximal number of steps Matt should jump.

 

Sample Input

21 10 552 10 336

 

Sample Output

Case #1: 2Case #2: 4

 

题意：在一条河中本来有n个石头，然后你可以随意添加石头，使得青蛙可以跳过河，但是青蛙每次会找单次最远的点跳，问青蛙最多跳多少次。

思路：贪心的讲，在跳的步数一样多的时候，肯定是此时青蛙的位置越小越好。用pos1和pos2记录青蛙上次跳跃的点和当前的点，如果前面L没有石头的话，就让他们每次跳L+1，这里是可以直接做除法算出来的。所以复杂度是o(n)的。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int val[200010];int main(){    int T,t,n,m,i,j,k,l,ans,pos1,pos2,point,ret;    scanf("%d",&T);    for(t=1;t<=T;t++)    {        scanf("%d%d%d",&n,&m,&l);        for(i=1;i<=n;i++)           scanf("%d",&val[i]);        if(n>0)        sort(val+1,val+1+n);        val[0]=0;        while(val[n]>=m && n>0)          n--;        val[n+1]=m;        n++;        ans=0;        pos1=-l-10;        pos2=0;        point=1;        while(pos2<m)        {            while(val[point]<=pos2 && point<n)               point++;            if(val[point]-pos2<=l)            {                while(val[point]-pos2<=l && point<=n)                   point++;                point--;                pos1=pos2;                pos2=val[point];                ans++;            }            else            {                k=(val[point]-pos2)/(l+1);                ans+=k*2;                if(val[point]-(pos1+(l+1)*k)>l)                {                    pos1=pos1+(l+1)*k;                    pos2=pos2+(l+1)*k;                }                else                {                    ret=pos1;                    pos1=pos2+(l+1)*(k-1);                    pos2=ret+(l+1)*k;                    ans--;                }            }        }        printf("Case #%d: %d\n",t,ans);    }}