HDOJ - 4004 The Frog's Games

The Frog's Games

Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u

Description

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).

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m. 
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.

Output

For each case, output a integer standing for the frog's ability at least they should have.

Sample Input

6 1 2
2
25 3 3
11 
2
18 

Sample Output

4
11

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大致题意：青蛙通过河里的石头跳到对岸，河长L，河里有石头N块，最多只能跳M次。每个石块到岸边的距离已给出，求青蛙跳的最短距离。

解题报告：很明显河里的石头距离岸边的长度是递增的，青蛙不可能来回跳！所以可用二分来处理。刚做的时候一脸懵逼，毫无头绪，看了别人的集体思路，一点点弄明白了。。如果青蛙连两块石头间的最远距离都跳不到，它是不可能到对岸的，所以左区间应从最远距离开始。

#include <cstdio>
#include <algorithm>
using namespace std;
int n,m,d[500001];
int sum(int mid){	
    int i,j,k=0,y=0;
    for(i=1;i<n+2;i++){
        y+=d[i]-d[i-1];//y代表从岸边跳到第i个石头上所用的距离  
        if(y>mid){//当y大于mid时，可以mid跳到前一个石头上 
			y=d[i]-d[i-1];//再次从第i个开始累计 
			k++;// 需要跳的次数  
		}
    }
    return y>mid?1<<30:k+1;//1<<30即2^30，如果y>mid，说明最后不能跳到对岸 
}
int main(){
    int l,left,right,i,j,k,mid,x,s;
    while(~scanf("%d%d%d",&l,&n,&m)){
        for( i=1;i<=n;i++)
        	scanf("%d",&d[i]);
        d[n+1]=l;//整条河的长度 
        d[0]=0;	
        sort(d+1,d+n+1);
        for(i=1,j=0;i<n+2;i++)
        	if(j<d[i]-d[i-1])
        		j=d[i]-d[i-1];//两点之间的最远距离
        left=j;right=l;//最远距离为左区间 
        while(left<right){//开始二分 
            mid=(left+right)/2;
            if(sum(mid)<=m)
				right=mid;// 右区间能取到mid!因为从左端起跳，临界状态 
            else
				left=mid+1;
        }
        printf("%d\n",left);
    }
    return 0;
}