POJ-2456Aggressive cows

来源：互联网发布：最新网络流行歌曲编辑：程序博客网时间：2024/05/21 14:52

题意：有N个坐标点，选择其中C个点。要使选择的点的相邻的距离的最小值最大。

思路：根据最小值的性质，枚举最小值。

c（x）表示距离的最小值距离为x时，能否找到满足题目的C个点。满足表示x可以更大。不满足，表示x应该更小。

#include <iostream>#include <map>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <queue>#include <stack>#include <functional>#include <set>#include<sstream>#include <cmath>using namespace std;#define pb push_back#define PB pop_back#define bk back()#define fs first#define se second#define INF 1001000#define sq(x) (x)*(x)#define eps (1e-10)#define clr(x) memset((x),0,sizeof (x))#define cp(a,b) memcpy((a),(b),sizeof (b))typedef long long ll;typedef unsigned long long ull;typedef pair<int,int> P;const int maxn=101000;int N,C;int cow[maxn];bool c(int m){    int num=1;    int last=cow[0];    for(int i=1;i<N;i++)    {        if(cow[i]-last>=m)        {            last=cow[i];            num++;        }    }    return num>=C;}int main(){    scanf("%d%d",&N,&C);    for(int i=0;i<N;i++) scanf("%d",&cow[i]);    sort(cow,cow+N);    int left,right,mid;    left=0,right=*max_element(cow,cow+N)+1;    while(right-left>1)    {        mid=(left+right)/2;        if(c(mid)) left=mid;        else       right=mid;    }    printf("%d\n",left);    return 0;}

0 0