Football Games HDU - 5873（兰道定理）

最新推荐文章于 2021-11-12 10:31:32 发布

原创最新推荐文章于 2021-11-12 10:31:32 发布 · 444 阅读

0 ·

CC 4.0 BY-SA版权

思维技巧专栏收录该内容

181 篇文章

订阅专栏

博客围绕足球比赛HDU - 5873题展开，比赛中每两支球队都要比一场，赢、输、平分别对应2、0、1分。题目给定球队比分情况，判断其是否合法。分析时起初用简单条件判断通过，后发现兰道定理是正解，可据此判断比分序列合法性。

Football Games HDU - 5873

A mysterious country will hold a football world championships—Abnormal Cup, attracting football teams and fans from all around the world. This country is so mysterious that none of the information of the games will be open to the public till the end of all the matches. And finally only the score of each team will be announced.

At the first phase of the championships, teams are divided into M
groups using the single round robin rule where one and only one game will be played between each pair of teams within each group. The winner of a game scores 2 points, the loser scores 0, when the game is tied both score 1 point. The schedule of these games are unknown, only the scores of each team in each group are available.

  When those games finished, some insider revealed that there were some false scores in some groups. This has aroused great concern among the pubic, so the the Association of Credit Management (ACM) asks you to judge which groups' scores must be false.

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

  For each case, the first line contains a positive integers M, which is the number of groups.

The i-th of the next M lines begins with a positive integer Bi representing the number of teams in the i-th group, followed by Bi
nonnegative integers representing the score of each team in this group.

number of test cases <= 10
M<= 100
B[i]<= 20000
score of each team <= 20000

Output
For each test case, output M
lines. Output F" (without quotes) if the scores in the i-th group must be false, outputT” (without quotes) otherwise. See samples for detail.
Sample Input

2
3 0 5 1
2 1 1

Sample Output

F
T

题意：

有B支足球队，每两支队伍都要比一场，赢了得2分，输了得0分，平局双方各得1分。

每次给定B个球队的比分情况问是否合法

分析：

当时看做出来的时候感觉不明觉历，随随便便找了几个条件，看是否满足，但是总感觉还差点什么结果就过了

之后搜题解大部分又都说是思维题模拟题，但是却发现兰道定理这个奇妙的东西，恰好就是这个问题的正解

兰道定理：

兰道定理又称竞赛图定理，是一个定义在有向图上的概念，顾名思义，它可以想象成n个人两两对决，赢得向输的连边，其实就是给一副完全图的无向边定了方向。

定义一个竞赛图的比分序列(score sequence)，是把竞赛图的每一个点的出度从小到大排列得到的序列。（这里相当于赢一句得1分）
一个长度为n的序列 $s = (s_1 \le s_2 \le ...\le s_n) ,n\ge1$ ，是合法的比分序列当且仅当：

\forall 1 \leq k \leq n, \sum i = 1 k s i \geq (k 2)

$\forall 1 \le k \le n,\sum_{i=1}^ks_i\ge \binom{k}{2}$

且 $k = n$ 时这个式子必须取等号

因此我们直接按照这个式子判断就行了，只不过这道题目赢一句得2分，因此给 $\binom{k}{2}$ 乘2就行了

code：

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 20010;
int a[MAXN];
int main(){
    int t;
    while(scanf("%d",&t) != EOF){
        while(t--){
            memset(a,0,sizeof(a));
            int n,i,sum = 0;
            bool flag = true;
            scanf("%d",&n);
            for(int i = 0; i < n; i++){
                scanf("%d",&a[i]);
                sum += a[i];
            }
            if(sum != n*(n-1)){
                flag = false;
                puts("F");
                continue;
            }
            sort(a,a+n);
            sum = 0;
            for(int i = 0; i < n-1; i++){
                sum += a[i];
                if(sum >= (i + 1) * i) continue;
                else{
                    flag = false;
                    puts("F");
                    break;
                }
            }
            if(flag) puts("T");
        }
    }
    return 0;
}