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 Di from the start (0 < Di < 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
Output
Sample Input
25 5 2 2 14 11 21 17
Sample Output
4
取得最大距离与最小距离，那么移除石头后的最短距离必然在之之间
那么我们可以再这个区间内对结果进行二分，看二分出的最短距离是否能符合移除m个石头的方案即可
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define up(i,x,y) for(i=x;i<=y;i++) #define down(i,x,y) for(i=x;i>=y;i--) #define mem(a,b) memset(a,b,sizeof(a)) #define w(a) while(a) #define ll long long int a[50005]; int main() { int l,n,m,i,j,k,minn,maxn,mid,sum,cnt; w(~scanf("%d%d%d",&l,&n,&m)) { minn=1000000005; maxn=l; a[0]=0,a[n+1]=l; up(i,1,n) scanf("%d",&a[i]); n++; sort(a,a+n); up(i,1,n) minn=min(minn,a[i]-a[i-1]); w(minn<=maxn) { mid=(maxn+minn)>>1; cnt=sum=0; up(i,1,n) { if((sum+=a[i]-a[i-1])<=mid) cnt++; else//连续的几个石头距离和大于mid的话，再把连续距离清0重新枚举 sum=0; } if(cnt<=m) minn=mid+1; else maxn=mid-1; } printf("%d\n",minn); } return 0; }
POJ3258：River Hopscotch(二分),布布扣,bubuko.com
原文地址：http://blog.csdn.net/libin56842/article/details/38660579
