http://acm.hdu.edu.cn/showproblem.php?pid=5046 n城市建k机场使得,是每个城市最近机场的距离的最大值最小化 二分+DLX 模板题 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #include <vector>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5046 题意:n个城市修建m个机场,使得每个城市到最近进场的最大值最小. 思路:二分+dlx搜索判定. #include <iostream> #include <map> #include <string> #include <stdio.h> #include <vector> #include <set> #include <a…

Fire station Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1308    Accepted Submission(s): 434 Problem Description A city's map can be seen as a two dimensional plane. There are N houses in…

HDU 5046 Airport 题目链接 题意:给定一些机场.要求选出K个机场,使得其它机场到其它机场的最大值最小 思路:二分+DLX反复覆盖去推断就可以 代码: #include <cstdio> #include <cstring> using namespace std; const int MAXNODE = 4005; const int MAXM = 65; const int MAXN = 65; const int INF = 0x3f3f3f3f; int K;…

转自:http://www.cnblogs.com/kuangbin/archive/2012/08/23/2653003.html   一种是直接根据公式计算的,另外一种是二分算出来的.两种方法速度都很快,充分体会到二分的效率之高啊~~~   题目中一个很重要的条件就是 (Lx*Lx+Ly*Ly) < vD*vD < vB*vB , 这样说明一定是可以追上的,而且可以以最大的距离射中,所以第一问的答案一定就是L的. 假设追击者跑的时间是 t1,那么肯定子弹飞行时间就是 L/vB 了 那么此时…

题目链接:http://poj.org/problem?id=3740 题意: 是否从0,1矩阵中选出若干行,使得新的矩阵每一列有且仅有一个1? 原矩阵N*M $ 1<= N <= 16 $ , $ 1 <= M <= 300$ 解法1:由于行数不多,二进制枚举选出的行数,时间复杂度为O((1<<16)*K), 其中K即为判断选出的行数是否存在相同的列中有重复的1: 优化:将1状压即可,这样300的列值,压缩在int的32位中,使得列数“好像”小于10了:这样每次只需要…

Easy Finding Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16178   Accepted: 4343 Description Given a M×N matrix A. Aij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.…

同样是二分+DLX即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define LL __int64 using namespace std; ; ; ; const int inf=0x3f3f3f3f; const double rinf=1e10; int L[maxn], R[maxn],…

模板 #1: int binarySearch(vector<int>& nums, int target){ if(nums.size() == 0) return -1; int left = 0, right = nums.size() - 1; while(left <= right){ // Prevent (left + right) overflow int mid = left + (right - left) / 2; if(nums[mid] == targe…

Description The country of jiuye composed by N cites. Each city can be viewed as a point in a two- dimensional plane with integer coordinates (x,y). The distance between city i and city j is defined by d ij = |x i - x j| + |y i - y j|. jiuye want to…