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LA 3644 X-Plosives

2014-12-25 18:08 190 查看
题意:给出n组元素装入一个箱中,有些元素不能混合,判断有多少元素不能装入箱中

链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=12648

思路:明显的并查集判断是否存在回路

注意点:合并操作又写错了。。查了好久,跪烂。。

以下为AC代码:

RunID
User

OJ

All

Prob ID

Result

All
Memory

(KB)
Time

(ms)
Language

All

Length

(Bytes)
Submit Time
3112802luminous11
UVALive
3644
Accepted06C++11 4.8.2
12442014-12-25 18:04:56
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;

int num[100005];

int find ( int x )
{
return x == num[x] ? x : num[x] = find( num[x] );
}

int main()
{
ios::sync_with_stdio( false );
int a, b;
while ( cin >> a )
{
for ( int i = 0; i < 100005; i ++ )
num[i] = i;
int k = 0;
int cnt = 0;
for ( ; a != -1; )
{
cin >> b;
a = find ( a );
b = find ( b );
if ( a == b )
{
cnt ++;
}
else
{
num[a] = b;
}
cin >> a;
}
cout << cnt << endl;
}
return 0;
}
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