Friends and Enemies HDU - 5874（思维）

最新推荐文章于 2019-05-23 20:50:42 发布

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博客围绕HDU - 5874题目展开，题目给定岛上m个人、n种石头，要求两人见面要么是朋友（项链至少一种石头颜色相同）要么是敌人（无相同颜色石头）。解题思路是将m个人分两个部落，部落内为朋友、部落间为敌人，按最坏情况判断所需石头种类，进而判断能否满足条件。

Friends and Enemies HDU - 5874

 On an isolated island, lived some dwarves. A king (not a dwarf) ruled the island and the seas nearby, there are abundant cobblestones of varying colors on the island. Every two dwarves on the island are either friends or enemies. One day, the king demanded that each dwarf on the island (not including the king himself, of course) wear a stone necklace according to the following rules:

  For any two dwarves, if they are friends, at least one of the stones from each of their necklaces are of the same color; and if they are enemies, any two stones from each of their necklaces should be of different colors. Note that a necklace can be empty.

  Now, given the population and the number of colors of stones on the island, you are going to judge if it's possible for each dwarf to prepare himself a necklace.

Input
Multiple test cases, process till end of the input.

  For each test case, the one and only line contains 2 positive integers M,N (M,N<231)
representing the total number of dwarves (not including the king) and the number of colors of stones on the island.

Output
For each test case, The one and only line of output should contain a character indicating if it is possible to finish the king’s assignment. Output T" (without quotes) if possible,F” (without quotes) otherwise.
Sample Input

20 100

Sample Output

题意：

说一个岛上有m个人，n种石头，两个人见面要么是朋友要么是敌人，是朋友的要求是两个矮人的身上的项链上至少有一个石头的颜色相同，是敌人的要求就是两个矮人的身上没有石头颜色相同。问给出m，n的值，能否满足使得两个人见面要么是敌人要么是朋友的条件。

思路：

1、其实题意隐晦成这个样子：就是让你将m个人分成两个部落，使得部落之内都是朋友，部落之间是敌人。

2、那么最坏的条件就是将m个人分成两个部落，每个部落m/2个人。那么满足两两之间见面都有关系，那么需要 $\frac{m}{2}\cdot \frac{m}{2}条边，也就是需要\frac{m^2}{4}$ 种石头。

按照上述判断一下即可。

code:

#include <bits/stdc++.h>
using namespace std;
int main(){
    long long n,m;
    while(~scanf("%lld%lld",&m,&n)){
        if(m * m / 4 > n) printf("F\n");
        else printf("T\n");
    }
    return 0;
}