首页 > ACM题库 > HDU-杭电 > hdu 1939 He is offside![解题报告]C++
2013
12-26

hdu 1939 He is offside![解题报告]C++

He is offside!

问题描述 :

Hemisphere Network is the largest television network in Tumbolia, a small country located east of South America (or south of East America). The most popular sport in Tumbolia, unsurprisingly, is soccer; many games are broadcast every week in Tumbolia.

Hemisphere Network receives many requests to replay dubious plays; usually, these happen when a player is deemed to be offside by the referee. An attacking player is offside if he is nearer to his opponents’ goal line than the second last opponent. A player is not offside if
●he is level with the second last opponent or
●he is level with the last two opponents.

Through the use of computer graphics technology, Hemisphere Network can take an image of the field and determine the distances of the players to the defending team’s goal line, but they still need a program that, given these distances, decides whether a player is offside.

输入:

The input file contains several test cases. The first line of each test case contains two integers A and D separated by a single space indicating, respectively, the number of attacking and defending players involved in the play (2 <= A,D <= 11). The next line contains A integers Bi separated by single spaces, indicating the distances of the attacking players to the goal line
(1 <= Bi <= 104). The next line contains D integers Cj separated by single spaces, indicating the distances of the defending players to the goal line (1 <= Cj <= 104). The end of input is indicated by A = D = 0.

输出:

The input file contains several test cases. The first line of each test case contains two integers A and D separated by a single space indicating, respectively, the number of attacking and defending players involved in the play (2 <= A,D <= 11). The next line contains A integers Bi separated by single spaces, indicating the distances of the attacking players to the goal line
(1 <= Bi <= 104). The next line contains D integers Cj separated by single spaces, indicating the distances of the defending players to the goal line (1 <= Cj <= 104). The end of input is indicated by A = D = 0.

样例输入:

2 3
500 700
700 500 500
2 2
200 400
200 1000
3 4
530 510 490
480 470 50 310
0 0

样例输出:

N
Y
N


题目判断越界的条件是, 攻击队员如果比倒数第二个防守队员距离防守线近的话,就是越界。 嘻嘻

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
 int a, d;
 int A[12], D[12];
 int i,j, k;
 while(cin>>a>>d)
 {
  if(a==d&&d==0) break;
  for(i=0;i<a;i++)
   cin>>A[i];
  for(i=0;i<d;i++)
   cin>>D[i];
  sort(A,A+a);
  sort(D,D+d);
  if(A[0]<D[1]) 
  {
   cout<<"Y\n";
   continue;
  }
  else
  {
   cout<<"N\n";
  }
 }
 return 0;
}

 


  1. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c

  2. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }

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