poj 1636 Prison rearrangement

Prison rearrangement

 

Time Limit: 3000MS		Memory Limit: 10000K
Total Submissions: 2158		Accepted: 971

Description

In order to lower the risk of riots and escape attempts, the boards of two nearby prisons of equal prisoner capacity, have decided to rearrange their prisoners among themselves. They want to exchange half of the prisoners of one prison, for half of the prisoners of the other. However, from the archived information of the prisoners' crime history, they know that some pairs of prisoners are dangerous to keep in the same prison, and that is why they are separated today, i.e. for every such pair of prisoners, one prisoners serves time in the first prison, and the other in the second one. The boards agree on the importance of keeping these pairs split between the prisons, which makes their rearrangement task a bit tricky. In fact, they soon find out that sometimes it is impossible to fulfil their wish of swapping half of the prisoners. Whenever this is the case, they have to settle for exchanging as close to one half of the prisoners as possible.

Input

On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two non-negative integers m and r, 1 < m < 200 being the number of prisoners in each of the two prisons, and r the number of dangerous pairs among the prisoners. Then follow r lines each containing a pair xi yi of integers in the range 1 to m,which means that prisoner xi of the first prison must not be placed in the same prison as prisoner yi of the second prison.

Output

For each test scenario, output one line containing the largest integer k <= m/2 , such that it is possible to exchange k prisoners of the first prison for k prisoners of the second prison without getting two prisoners of any dangerous pair in the same prison.

Sample Input

3
101 0
3 3
1 2
1 3
1 1
8 12
1 1
1 2
1 3
1 4
2 5
3 5
4 5
5 5
6 6
7 6
8 7
8 8

Sample Output

50
0
3
原文地址
题目意思：两个监狱，各有n个犯人，每个两个监狱之间一些犯人之间有一定的关系，对于有关系的犯人不能放在同一个监狱，原状态肯定是满足的，
         因为存在这种关系的不存在同一个监狱的。求最大交换次数使得条件依然满足，并且交换次数不能超过n/2。

后记：我们两队的比赛题，琨哥说题目很水，我信了，这能水，唉，高估我们能力了，主要是用到 dfs + o1 背包问题，但是的确很难想到，我还想着用并查集呢，不会，参考了下别人的代码；

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int SIZE = ;
 int m;
 int r;
 //map[i][j]表示i和j是否会冲突
 int map[SIZE][SIZE];
 //A组里的人数
 int aSize;
 //b组里的人数
 int bSize;
 //dp[i][j] 表示用A组的i个人换B组的j个人是否可行
 bool dp[SIZE][SIZE];
 //visited[0][i] 表示用A组中的点i是否被访问过
 //visited[1][i] 表示用B组中的点i是否被访问过
 bool visited[][SIZE];
 void Init()
 {
     memset( map, , sizeof(map) );
     memset( visited, , sizeof(visited) );
     memset( dp, , sizeof(dp) );
 }
 void Input()
 {
     cin >> m >> r;
     for( int i=; i<r; i++ )
     {
         int a, b;
         cin >> a >> b;
         map[a][b] = ;
     }
 }
 //side=0  表示当前正在搜索A组
 //side=1  表示当前正在搜索B组
 //id 表示当前正在搜索的编号
 void DFS( int side, int id )
 {
     visited[side][id] = true;
     //如果当前搜索的是A组
     if( side ==  )
     {
         //记录A组中的元素个数
         aSize++;
         for( int i=; i<=m; i++ )
         {
             //搜索的是B组中对应的点
             if( map[id][i] && !visited[][i] )//看一组的 id ，是否有和二组的 i ，相连的不，并且二组的没有被标记；
             {
                 DFS( , i ); //搜寻一组中是否有与二组相连的数；
             }
         }
     }
     else
     {
         bSize++;
         for( int j=; j<=m; j++ )
         {
             if( map[j][id] && !visited[][j] )
             {
                 DFS( , j );
             }
         }
     }
 }
 //利用二维背包计算
 void Knapsack()
 {
     dp[][] = true;
     for( int x=m/; x>=aSize-; x-- )
     {
         for( int y=m/; y>=bSize-; y-- )
         {
             if( dp[x][y] || dp[x - aSize][y - bSize] )
             {
                // printf("%d %d\n",x,y);
                 dp[x][y] = true;
             }
         }
     }
 }
 void Output()
 {
     for( int i=m/; i>=; i-- )
     {
         if( dp[i][i] )                         // dp[i][i]; 表示各方都拿出来 i 个人，进行交换；
         {
             cout << i << endl;
             break;
         }
     }
 }
 int main()
 {
     int caseNum;
     cin >> caseNum;
     while( caseNum-- )
     {
         Init();
         Input();
         for( int i=; i<=m; i++ )
         {
             //跳过已经处理过的节点
             if( visited[][i] ) continue;
             //计算A、B中的人数
             aSize = ;
             bSize = ;
             DFS( , i );    //  搜索一次，就出现两组不相容的团体；
             //利用二维背包计算
             Knapsack();
         }
         for( int i=; i<=m; i++ )
         {
             if( visited[][i] ) continue;
             aSize = ;
             bSize = ;
             DFS( , i );
             Knapsack();
         }
         Output();
     }
     return ;
 }