基础回顾--二分
二分求解问题类型:有界有序数据区间上的一个最优解
关键点:1)确定上下界
2)解存在的区间为【left,right】,二分时的mid为当前的猜想解,最好用一个变量ans存储每次最优解,以免最终输出的混乱
3)注意终止的范围while(left <= right)
复杂度:logN
栗子:P1316 丢瓶盖
/* 有界最优解问题 ==> 二分枚举 */ #include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int maxn = 100000 + 4; int a[maxn]; int n, m, ans; int x1=0x3f3f3f3f, x2; inline int max(int a, int b) { return a>b? a:b; } inline int min(int a, int b) { return a<b? a:b; } inline bool check(int q) { int pre = a[1], cur = 0; int cnt = 0; for(int i=2; i<=n; ++i) { cur = a[i]; if(cur - pre < q) continue; else { pre = a[i]; cnt += 1; } } return cnt >= m-1; } int main() { scanf("%d %d", &n, &m); for(int i=1; i<=n; ++i) scanf("%d", a+i); sort(a+1, a+1+n); x2 = a[n] - a[1]; for(int i=2; i<=n; ++i){ x1 = min(x1, a[i]-a[i-1]); } while(x1 <= x2) { // mid为当前猜想的解 int mid = x1 + (x2-x1)/2; if(check(mid)) { //定义一个变量存储最优解 ans = max(mid, ans); x1 = mid+1; } else x2 = mid-1; } printf("%d\n", ans); return 0; }