POJ 3258 River Hopscotch [NOIP2015 D2T1] (洛谷 P2678 跳石头)
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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 Mrocks (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 ofM rocks.
Input
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
Sample Input
25 5 2214112117
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).
(翻译见洛谷 P2678~)
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二分。
洛谷上的评级是 提高- 。。。对我的智商产生了怀疑。。。这题真的有这么简单么?
(用50的例子想)
初始的时候lef=0,rit=l,
然后二分,中间值移的石头大于m(即举得过大)时,rit=k-1,
其他的时候lef=k+1,更新ans值.
//3258/*怎样移可以使最小距离最大*/ #include<cstdio>#include<cstring>#include<algorithm>using namespace std;int l,n,m,k,a[50006],ans;int move(int u){int as=0,sum=0;for(int i=1;i<=n+1;i++){if(a[i]-a[as]<u) sum++;else as=i;}return sum;}int main(){scanf("%d%d%d",&l,&n,&m);for(int i=1;i<=n;i++) scanf("%d",&a[i]);a[0]=0;a[n+1]=l;sort(a+1,a+n+1);int lef=0,rit=l;while(rit>=lef){k=(rit+lef)/2;if(move(k)>m) rit=k-1;else{lef=k+1;ans=k;}}printf("%d",ans);}
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