hdu 5037 Frog

代码转载至http://blog.csdn.net/zz_1215/article/details/39453753

说说我的理解。。。

用now表示当前所在点

pre表示前一步所在点

next表示下一步要去的点

每（l+1）次跳两步肯定是最优的解，但是要处理一下两种情况

第一种就是pre+（l+1）（next-now）/（l+1）表示now到达了next之前最近的（l+1）的倍数的距离时pre的位置

如果这时候的pre的位置和next的位置之差小于等于l，那么就能直接到达next，所以就让now=next

第二种是pre和next的距离大于l，那么就说明还需要走一步才能到达next

所以让pre=now（走了l+1的倍数之后的now），now=next，同时ans++；

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
#include<ctime>
using namespace std;
typedef long long i64;
#define SS(a) scanf("%d",&a)
#define MM(name,what) memset(name,what,sizeof(name))
#define MC(a,b) memcpy(a,b,sizeof(b))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);

int n, m, l;
vector<int>v;
int id;

int get(){
return v[id++];
}

int start(){

int now = 0;
int ans = 0;
int temp;
v.push_back(m);
int pre = -l;
for (int i = 0;i < v.size(); i++){
int next = v[i];
temp = next - now;
int t2 = temp / (l + 1);
pre += t2*(l + 1);
ans += t2 * 2;
if (next - pre>l){
pre = now + t2*(l + 1);
now = next;
ans++;
}
else if (next - pre <= l){
now = next;
}
}
return ans;
}

int main(){
int T;
cin >> T;
for (int tt = 1; tt <= T;tt++){
cin >> n >> m >> l;
v.clear();
int now;
for (int i = 1; i <= n; i++){
//cin >> now;
SS(now);
v.push_back(now);
}
sort(v.begin(), v.end());
cout << "Case #" << tt << ": ";
cout << start() << endl;
}
return 0;
}