BZOJ4525——[Usaco2016 Jan]Angry Cows

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1、题意：给一堆可以的限制长度的区间。。。区间的长度是你控制的。。。但是只有一个长度。。。求最短长度覆盖所有的点
2、分析：发现可以二分。。。那二分吧。。。。。然后我们从头向后扫一遍直接判断能否直接覆盖。。。然后就可以AC了《大赛后一水系列》

#include <map>#include <set>#include <queue>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define LL long long#define inf 2147483647#define M 100010inline int read(){    char ch = getchar(); int x = 0, f = 1;    while(ch < '0' || ch > '9'){        if(ch == '-') f = -1;        ch = getchar();    }    while('0' <= ch && ch <= '9'){        x = x * 10 + ch - '0';        ch = getchar();    }    return x * f;} int a[M];int n, K;inline bool check(int x){    int l = 1, o = 0;    for(int i = 1; i <= n; i ++){        if(a[i] - a[l] > x * 2){            o ++;            l = i;        }     }    o ++;    return o <= K;}int main(){    n = read(), K = read();    for(int i = 1; i <= n; i ++) a[i] = read();    sort(a + 1, a + n + 1);    int l = 0, r = 1000000000, t = 1000000000;    while(l <= r){        int mid = (l + r) / 2;        if(check(mid)) r = (t = mid) - 1;        else l = mid + 1;    }    printf("%d\n", t);    return 0;}

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