poj1029 模拟

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如题：http://poj.org/problem?id=1029

False coin

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17541 Accepted: 4898

Description

The "Gold Bar"bank received information from reliable sources that in their last group of N coins exactly one coin is false and differs in weight from other coins (while all other coins are equal in weight). After the economic crisis they have only a simple balance available (like one in the picture). Using this balance, one is able to determine if the weight of objects in the left pan is less than, greater than, or equal to the weight of objects in the right pan.
In order to detect the false coin the bank employees numbered all coins by the integers from 1 to N, thus assigning each coin a unique integer identifier. After that they began to weight various groups of coins by placing equal numbers of coins in the left pan and in the right pan. The identifiers of coins and the results of the weightings were carefully recorded.
You are to write a program that will help the bank employees to determine the identifier of the false coin using the results of these weightings.

Input

The first line of the input file contains two integers N and K, separated by spaces, where N is the number of coins (2<=N<=1000 ) and K is the number of weightings fulfilled (1<=K<=100). The following 2K lines describe all weightings. Two consecutive lines describe each weighting. The first of them starts with a number Pi (1<=Pi<=N/2), representing the number of coins placed in the left and in the right pans, followed by Pi identifiers of coins placed in the left pan and Pi identifiers of coins placed in the right pan. All numbers are separated by spaces. The second line contains one of the following characters: '<', '>', or '='. It represents the result of the weighting:
'<' means that the weight of coins in the left pan is less than the weight of coins in the right pan,
'>' means that the weight of coins in the left pan is greater than the weight of coins in the right pan,
'=' means that the weight of coins in the left pan is equal to the weight of coins in the right pan.

Output

Write to the output file the identifier of the false coin or 0, if it cannot be found by the results of the given weightings.

Sample Input

5 32 1 2 3 4<1 1 4=1 2 5=

Sample Output

Source

Northeastern Europe 1998

思路：和poj1013的思路一模一样。

我的想法是去将所有能确定的真的确定下来，剩下的如果只剩一个一定是假的，否则多个可能是假的。

首先,=号两边的一定是真的。

考虑'>',左比右重，左边的heavy标记置为1，再去看它的light标记，如果是1，它是真的，因为一个假硬币不可能又重又轻，所以它真。

'<'同上。最后遍历一边True数组肯哪些为1就行了。

代码看着长，其实一看就懂。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int True[1005];
int heavy[1005];
int light[1005];

int main()
{
// freopen("C:\\1.txt","r",stdin);
int n,k;
cin>>n>>k;
int i,j;
for(i=0;i<k;i++)
{
  int p;
  cin>>p;
  int l[505]={0};
  int r[505]={0};
  for(j=1;j<=p;j++)
   scanf("%d",&l[j]);
  for(j=1;j<=p;j++)
   scanf("%d",&r[j]);
  char flg;
  getchar();
  flg=getchar();
  if(flg=='<')
  {
   int vis[1005]={0};
   for(j=1;j<=p;j++)
   {
    light[l[j]]=1;
    if(heavy[l[j]])
     True[l[j]]=1;
    vis[l[j]]=1;
   }
   for(j=1;j<=p;j++)
   {
    heavy[r[j]]=1;
    if(light[r[j]])
     True[r[j]]=1;
    vis[r[j]]=1;
   }
   for(j=1;j<=n;j++)
    if(!vis[j])
     True[j]=1;
  }
  else if(flg=='>')
  {
   int vis[1005]={0};
   for(j=1;j<=p;j++)
   {
    heavy[l[j]]=1;
    if(light[l[j]])
     True[l[j]]=1;
    vis[l[j]]=1;
   }
   for(j=1;j<=p;j++)
   {
    light[r[j]]=1;
    if(heavy[r[j]])
     True[r[j]]=1;
    vis[r[j]]=1;
   }
   for(j=1;j<=n;j++)
    if(!vis[j])
     True[j]=1;
  }
  else
  {
   for(j=1;j<=p;j++)
    True[l[j]]=1;
   for(j=1;j<=p;j++)
    True[r[j]]=1;
  }
}
int cnt=0;
int False;
for(i=1;i<=n;i++)
{
  if(True[i]==0)
  {
   cnt++;
   False=i;
  }
}
if(cnt==1)
  printf("%d\n",False);
else
  printf("0\n");
return 0;
}

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