hdu 5037 Frog(贪心)

来源：互联网发布：java中private的作用编辑：程序博客网时间：2024/05/16 01:34

题意是一个数轴上有一些石头，有个青蛙要从左边跳到右边，他一次最多能跳L步，然后他会尽可能少跳，你是上帝可以通过加石头来使他多跳，问最多能让青蛙跳几步。

是一个贪心的方法，需要记录该步前面一步的位置和当前位置，然后如果当前位置到当前位置加上L范围内有石头，那么直接跳到能跳的最远处，反之，放一个石头在前面一步位置加上L+1，当然，可能下一个石头的位置非常远而L又比较小，所以需要优化，需要推出公式，我就这里被坑到了。。直到比赛后才找到错误。。还有看别人写的好简单。。我写的好复杂。。真是智商的差距么。。好伤心

AC代码：

#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>#include<ctime>#include<string.h>#include<string>using namespace std;#define ll __int64#define eps 1e-10#define MOD 10000007template<class T>inline void scan_d(T &ret){    char c;    ret=0;    while((c=getchar())<'0'||c>'9');    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();}ll a[200005];int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o4.txt","w",stdout);#endif // GLQ    int t,i,n,cas=1;;    scanf("%d",&t);    while(t--)    {        ll m,l;        scanf("%d%I64d%I64d",&n,&m,&l);        for(i = 0; i < n; i++)            scan_d(a[i]);        sort(a,a+n);        ll pre,pos;        a[n] = m;        for(i = 0; i <= n && a[i] <= l; i++);        if(i == 0)        {            pre = 0;            pos = 1;        }        else        {            pre = 0;            pos = a[i-1];        }        ll ans = 1;        while(1)        {            if(pos+l >= m)            {                if(pos < m) ans++;                break;            }            int flag = 0;            while(pos+l >= a[i] && i < n)            {                flag = 1;                i++;            }            if(flag) i--;            if(pos+l < a[i] || i >= n)            {                ll temp = pre;//                cout<<"a:"<<a[i]<<" "<<pre<<" "<<pos<<" "<<ans<<endl;                ll t1 = (a[i]-pre)/(l+(ll)1);                pre = pos+(t1-(ll)1)*(l+(ll)1);//                cout<<pre<<endl;                pos = temp+t1*(l+(ll)1);                ans += 2*t1-(ll)1;                while(pre+l>=a[i])//已经可以到达a[i]了                {                    pos = a[i];                    i++;                }//                cout<<a[i]<<" "<<pre<<" "<<pos<<" "<<ans<<endl;                if(pre+l>=m)                {                    break;                }            }            else            {//                cout<<pre<<" "<<pos<<" "<<pos-pre<<" "<<a[i]<<" "<<a[i-1]<<" "<<ans<<endl;                pre = pos;                pos = a[i];                ans++;                i++;            }//            cout<<ans<<" "<<pre<<" "<<pos<<endl;        }        printf("Case #%d: %I64d\n",cas++,ans);    }    return 0;}

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