Football Games HDU - 5873

异常杯足球赛成绩校验

最新推荐文章于 2019-09-24 21:12:50 发布

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一场名为“异常杯”的神秘国际足球锦标赛吸引了全球的关注。本文介绍了一种算法，用于判断该赛事中各小组的成绩是否可能被篡改。通过单循环赛制下各队得分的特性，设计了一个简洁高效的解决方案。

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 MM 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 MM, which is the number of groups.
The ii-th of the next MM lines begins with a positive integer BiBi representing the number of teams in the ii-th group, followed by BiBi nonnegative integers representing the score of each team in this group.

number of test cases <= 10
M<= 100
Bii<= 20000
score of each team <= 20000
Output
For each test case, output MM 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

思路：233一开始想太简单了。。。本来足球是胜3平1负0啊这里胜2这样不管是什么结果，只要队伍数量定了那么他们的得分总和就是定的。考虑前k个队，这k个队必定会每个队之间都进行一次比赛，如果只考虑k个队之间的关系，那么得分应该是k*（k-1）但是还有多余的队伍，所以他们的得分必定是大于等于k (k-1)的这样我们把得分从小到大排序每次判断一下，最后再判断总和是不是n (n-1)

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<map>
using namespace std;

//thanks to pyf ...

#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
typedef pair<int, int> PII;
typedef long long ll;

const int N = 1e6 + 5;

int a[N];
int main()
{
    int T;
    while (scanf("%d", &T) != EOF)
    {
        while (T--)
        {
            int n;
            scanf("%d", &n);
            for (int i = 1; i <= n; i++)
                scanf("%d", a + i);
            sort(a + 1, a + 1 + n);
            int sum = a[1];
            int flag = 1;
            for (int i = 2; i <= n; i++)
            {
                sum += a[i];
                if (sum < i * (i - 1))
                    flag = 0;
            }
            if (!flag || sum != n * (n - 1))
                printf("F\n");
            else
                printf("T\n");
        }
    }
}