HDU 5873 Football Games 【】

Football Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617    Accepted Submission(s): 228

Problem Description

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, output ``T" (without quotes) otherwise. See samples for detail.

 

Sample Input

23 0 5 12 1 1

 

Sample Output

FT

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

              首先，一共C(m,2) 即m*(m-1)/2局比赛，一共m*(m-1)分

             不能同时出现两个0，最高分不能超过2*(m-1)

   

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))typedef long long LL;const int INF=0x7ffffff;const int M=2e4+1;int flag[M];int num[M];int i,j,k,n,m;int main(){    int M;    while(~scanf("%d",&M))    {        while(M--){          int n;          MS(flag,0);          scanf("%d",&n);          int sum=0;          int sum1=0;          rep(i,0,n){            scanf("%d",&flag[i]);            if(flag[i]==0)sum1++;            sum+=flag[i];          }          if(sum!=n*(n-1)||sum1>1){printf("F\n");continue;}          if(n==2){printf("T\n");continue;}          int mark=1;          sort(flag,flag+n);          int k=1;          int ans=0;          if(flag[n-1]>2*(n-1))mark=0;          if(mark)printf("T\n");          else printf("F\n");        }    }    return 0;}

  后来看了下别人怎么做的，感觉这个解法好神奇 ：

    假设我们将队伍编号为1-n，小号队总能打赢大号队，所以最后得分2*（n-1)，2*（n-2)。。。0

将实际得分按从大到小排序，用假设的得分减去实际得分，sum记录，如果小于0，退出，输出F，否则，说明该球队输过或平局过，这样多出的分就可以加在其他队上，最后sum值应为为0。

#include <bits/stdc++.h>using namespace std;int a[1000000];int main(){    int t;    while(~scanf("%d",&t))    {        while(t--)        {            int n;            scanf("%d",&n);            for(int i=1;i<=n;i++)                scanf("%d",&a[i]);                int flag=1;                int sum=0;                int he=2*(n-1);                sort(a+1,a+1+n);                reverse(a+1,a+1+n);                for(int i=1;i<=n;i++)                {                    sum+=(he-a[i]);                    if(sum<0)                    {                        flag=0;                        break;                    }                    he-=2;                }                if(sum==0&&flag)                    printf("T\n");                else                    printf("F\n");        }    }}