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  • hdu 4004 The Frog's Games

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4004

    The annual Games in frogs' kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
    are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog's longest jump distance). 
     
    题意:一条长L的河中有n块石头,青蛙在原位置通过最多跳m次(每次必须落在石头上或岸上)来通过这条河,求青蛙最远弹跳能力的最小值。
    解法:一般看到要求最大值最小什么的,会马上联想到二分搜索。此题就可以用二分来搞。
         二分青蛙最远弹跳能力的值,通过判断最多跳m次是否能够过河来判断此解是否有效,然后求最小即可。
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<cmath>
     6 #include<algorithm>
     7 #define inf 0x7fffffff
     8 using namespace std;
     9 const int maxn=500000+100;
    10 int an[maxn];
    11 int L,n,m;
    12 int isok(int len)
    13 {
    14     int step=0;
    15     int from=0;
    16     int i=1;
    17     while (from<L && i<=n+1)
    18     {
    19         int j=i;
    20         while (an[i]-from<=len && i<=n+1) {j=i ;i++ ; }
    21         step ++ ;
    22         from=an[j];
    23         i=j;
    24         if (step>m) return 0;
    25     }
    26     return 1;
    27 }
    28 int main()
    29 {
    30     while (scanf("%d%d%d",&L,&n,&m)!=EOF)
    31     {
    32         for (int i=1 ;i<=n ;i++) scanf("%d",&an[i]);
    33         an[n+1]=L;
    34         sort(an+1,an+n+1+1);
    35         int l=0,r=inf;
    36         int ans=0;
    37         while (l<=r)
    38         {
    39             int mid=(l+r)>>1;
    40             if (isok(mid)) {ans=mid ;r=mid-1 ; }
    41             else l=mid+1 ;
    42         }
    43         printf("%d
    ",ans);
    44     }
    45     return 0;
    46 }
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  • 原文地址:https://www.cnblogs.com/huangxf/p/4127157.html
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