poj3258——二分优化
poj3258——二分优化
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 8201 | Accepted: 3531 |
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
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 2 2 14 11 21 17
Sample Output
4
Hint
/** * start at 16:40 * end at 17:21 * time: 41 min * problem: poj3258 * */ #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; typedef long long ll; const int maxn=1000100; const ll INF=(1LL<<60); ll L; int N,M; ll pos[maxn]; //int times=1; bool judge(ll limit) { //cout<<times++<<endl; int del_cnt=0; for(int i=0,j=1;j<=N+1;j++){ ///用i和j指针控制贪心 if(pos[j]-pos[i]<limit) del_cnt++; else i=j; } if(del_cnt<=M) return true; return false; } ll BinSearch(ll low,ll high) { while(low<high){ ll mid=(low+high)/2+1;///下面是low=mid,所以这里+1,如果下面是high=mid,则不用+1。因为(2+3)/2=2,而不是等于3或2.5,,此处的处理避免死循环 //cout<<low<<" "<<high<<" "<<mid<<endl; if(judge(mid)) low=mid; else high=mid-1; } return low; } int main() { while(scanf("%lld%d%d",&L,&N,&M)!=EOF){ pos[0]=0; pos[N+1]=L; for(int i=1;i<=N;i++) scanf("%lld",&pos[i]); sort(pos,pos+N+1); //for(int i=0;i<=N+1;i++) cout<<i<<" "<<pos[i]<<endl; ll Min=INF; for(int i=1;i<=N+1;i++){ if(pos[i]-pos[i-1]<Min) Min=pos[i]-pos[i-1]; } //cout<<Min<<endl; printf("%lld\n",BinSearch(Min,L)); } return 0; }
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