POJ3258 River Hopscotch —— 二分

题目链接：http://poj.org/problem?id=3258

River Hopscotch

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15753		Accepted: 6649

Description

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance D_i from the start (0 < D_i < L).

To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).

FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

Input

Line 1: Three space-separated integers: L, N, and M 
Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

Output

Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input

25 5 2
2
14
11
21
17

Sample Output

4

Hint

Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

Source

USACO 2006 December Silver

 

 

题解：

 1.二分答案， 即最短距离。

2.枚举每一块石头（已经加入了起点和终点， 从第二块石头开始枚举）。 如果当前石头与前面一块石头的距离小于最短距离，则把当前石头移走；如果大于等于，则不进行任何操作。当操作完最后一块石头后（终点），如果移除的石头块数小于等于M， 则增大最短距离， 否则缩小最短距离。

3.问：题目要求不能移除起点和终点，从第二块石头开始枚举可以保证起点不被移除，但却不能保证最后一块石头不被移除，那为什么上述方法可行呢？

答：当枚举到终点时， 如果终点与前面一块石头的距离小于最短距离，按照代码的意思，是移除终点，但是实际操作我们可以移除终点前面的那块石头，使得终点与上上块石头的距离大于最短距离（必定如此，因为上上块石头与上块石头的距离就已经大于等于最短距离了）。那假如终点前面那块石头就是起点呢？这种情况不存在，因为这样的话，两点的距离就是L了，不可能出现L小于最短距离的情况。

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e5;

 int L, N, M;
 int a[MAXN];

 bool test(int mid)
 {
     int cnt = , pre = ;
     for(int i = ; i<=N; i++)
     {
         if(a[i]-a[pre]<mid) cnt++;
         else pre = i;
     }
     return cnt<=M;
 }

 int main()
 {
     while(scanf("%d%d%d",&L, &N, &M)!=EOF)
     {
         for(int i = ; i<=N; i++)
             scanf("%d",&a[i]);

         a[++N] = ; a[++N] = L;
         sort(a+, a++N);

         int l = , r = L;
         while(l<=r)
         {
             int mid = (l+r)>>;
             if(test(mid))
                 l = mid + ;
             else
                 r = mid - ;
             printf("%d %d\n", l, r);
         }
         printf("%d\n", r);
     }
 }